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WorksheetsTĩnh học: Hệ lực và Liên kết (Câu 1-12)
Total questions: 118
Worksheet time: 59mins
Hệ lực cân bằng là:
Hệ lực tương đương không
Các lực của hệ lực bằng nhau
Các lực của hệ lực ngược chiều nhau
Véc tơ chính của hệ lực bằng không
Hai hệ lực được gọi là tương đương khi:
Chúng có cùng tác dụng cơ học như nhau đối với một vật rắn
Chúng có giá trị bằng nhau
Chúng cùng đường tác dụng
Chúng làm vật rắn cân bằng
Hợp của hệ lực là:
Một lực duy nhất tương đương với hệ lực đã cho
Một hệ gồm hai lực cân bằng
Một ngẫu lực
Một lực và một ngẫu lực
Tác dụng của một lực lên vật rắn không đổi nếu:
Ta trượt dọc lực theo đường tác dụng của nó
Ta thêm vào đó một lực trực đối với lực đã cho
Ta thêm vào một ngẫu lực
Ta dời song song lực đã cho tới điểm khác
Liên kết là:
Những điều kiện cản trở chuyển động của vật
Sự chuyển động tương đối giữa hai vật tiếp xúc với nhau
Sự kẹp chặt vật thể bởi một cái ngàm
Những sợi dây treo vật
Phản lực liên kết là:
Lực do vật gây lên liên kết tác dụng lên vật khảo sát
Lực do vật khác sát tác dụng lên vật gây liên kết
Lực do hai vật tiếp xúc nhau
Lực ngược chiều với ngoại lực
Hệ lực đồng quy là:
Hệ lực có giá của các lực thành phần cắt nhau tại một điểm
Hệ lực có giá của các lực song song với nhau
Hệ lực có giá của các lực hợp với nhau một góc bất kỳ
Hệ lực có véc tơ chính của nó bằng không
Hệ lực song song là:
Hệ lực có giá của các lực thành phần song song với nhau
Hệ lực có giá của các lực cắt nhau tại một điểm
Hệ lực có giá của các lực hợp với nhau một góc bất kỳ
Hệ lực có véc tơ chính của nó bằng không
Ngẫu lực là:
Hệ gồm hai lực song song, khác giá, trái chiều và có trị số bằng nhau
Hệ gồm hai lực song song, cùng giá, ngược chiều và có trị số bằng nhau
Hệ gồm hai lực song song, cùng giá, ngược chiều và có trị số đối nhau
Hệ gồm hai lực song song, cùng giá, cùng chiều và có trị số bằng nhau
Hai ngẫu lực tương đương nhau là:
Hai ngẫu lực có véc tơ mô men ngẫu lực bằng nhau
Hai ngẫu lực có véc tơ mô men ngẫu lực ngược nhau
Hai ngẫu lực có véc tơ mô men ngẫu lực cùng độ lớn nhưng ngược chiều
Hai ngẫu lực có véc tơ mô men ngẫu lực song song với nhau
Trong hình vẽ, lực tác dụng trên đoạn AB là loại lực gì?
Hệ lực phân bố đều
Hệ lực phân bố bậc nhất
Lực tập trung
Hệ lực phân bố bậc hai
Trong hình vẽ, lực tác dụng tại điểm B là loại lực gì?
Lực tập trung
Mô men tập trung
Áp lực phân bố
Lực quán tính
A simply supported beam has a uniform distributed load of 5 kN/m on segment AB (2 m) and a concentrated load of 10 kN at point C, 1 m beyond B. Which classification best describes the loading on segment AB?
Uniformly distributed load over segment
Concentrated force at a point on segment
Linearly varying distributed load over segment
Couple moment acting on the segment
On a beam AB, the distributed load increases linearly from 4 kN/m at A to 8 kN/m at B across span AC (3 m). What is the correct classification of the load acting on segment AC?
Linearly varying distributed load over length
Uniformly distributed load over length
Concentrated forces at equal spacing
System of concurrent point forces
On the same beam, consider segment CB (3 m) where the distributed load continues increasing linearly from its value at C to 8 kN/m at B. Which classification best fits the load on segment CB?
Linearly varying distributed load over length
Uniformly distributed load over length
Couple of equal and opposite forces
Parallel system of equal point loads
A cylindrical pin at support B permits rotation but prevents translation in the plane. Which type of connection does this represent?
Pinned (hinge) support connection
Fixed (built-in) connection
Frictionless roller support
Knife-edge support connection
In Figure 1.1, what is the support at A of the beam AB under a uniform load p = qa?
Fixed support (encastre)
Pinned support (fixed hinge)
Rigid bar connection
Roller support (movable)
Cable support (flexible)
In Figure 1.1, what is the support at C of the beam when the right end allows horizontal movement but resists vertical motion?
Roller support (movable)
Pinned support (fixed hinge)
Fixed support (encastre)
Clamped bracket connection
Two-force member joint
In Figure 1.2, identify the support at A where the beam end is rigidly embedded into a wall and resists rotation.
Fixed support (encastre)
Pinned support (hinged)
Roller support (movable)
Cable support (flexible)
Pin–slot connection
For segment AB in the diagram with 5 kN/m uniformly distributed over AB and a 10 kN point load at C, what type of loading acts on AB?
Uniformly distributed load
Linearly varying distribution
Concentrated force only
Distributed moment only
Pair of equal point loads
In the final diagram, which type of support is shown at B if it prevents rotation and translation in all directions?
Fixed support (encastre)
Pinned support (hinged)
Roller support (movable)
Elastic spring support
Cable support (flexible)
In the diagram, a straight bar CD is connected at point C to a fixed corner and supports a small roller at D. A cable from the upper left pulls at C, and a horizontal member CB with a roller at B carries a vertical load P. For the bar CD, what is the direction of its reaction force at joint C?
Along the line connecting C and D
Perpendicular to member CD
Collinear with member CB
No reaction force is present
In the L‑shaped frame shown, a vertical member AB is fixed at A, a horizontal member CD is attached by a pin at C, and the lower right corner E sits on a small roller on a horizontal track. What type of support exists at point E?
Movable (roller) support
Fixed (built‑in) support
Clamped (cantilever) support
Two‑force link support
Consider a beam AB resting on supports with a vertical link at C connected to an inclined bar CD that is clamped to a wall at D. At point C, the beam connects to the inclined bar by a pin. What type of connection is at C?
Fixed (built‑in) support
Movable (roller) support
Pinned (hinged) joint
Rigid link connection
In the frame shown, bar BD forms a 60° angle with the horizontal x-axis. If the reaction at B on BD is along the member AB which is horizontal, what is the angle between the reaction of bar BD and the x-axis?
60 degrees
30 degrees
45 degrees
90 degrees
For the same frame, determine the angle between the reaction of bar BD and the vertical y-axis when BD is inclined at 60° to the x-axis.
30 degrees
60 degrees
45 degrees
90 degrees
In the L-shaped member supported at point E by a slot along the horizontal line AE, what is the direction of the reaction at E? Assume the slot constrains motion only perpendicular to AE.
Vertical upward
Horizontal to the right
At 45° to the horizontal
Parallel to segment CD
A beam is supported at A and C with a uniformly distributed load of 5 kN/m on the left span and a 10 kN point load near C. At A, the support prevents both translation and rotation. What type of connection is at A?
Fixed support (built-in)
Roller support (movable)
Pin support (hinged)
Cable support (tension)
In the figure, joint A attaches a beam to a wall with no rotation allowed and all translations restrained. Which support best describes joint A?
Fixed support (clamped) connection
Pinned hinge allowing rotation
Movable roller support only
Cable or tie transmitting tension
A shaft passes through a bracket at A using a pin that allows rotation but prevents translation of the pin center. Which connection type is at A?
Pinned hinge (revolute joint)
Two-force straight bar link
Movable roller bearing support
Rigid built-in fixed support
A simply supported beam rests on a support at A that prevents vertical translation but allows horizontal movement and rotation. Which support is modeled at A?
Fixed pin support at the base
Movable roller (simple support)
Cable support in pure tension
Rigid clamp with no rotation
Location B uses a support that restrains both horizontal and vertical translation while allowing rotation about the support. Which designation best matches support B?
Pinned support (hinge) at the base
Movable roller allowing sliding
Clamped fixed built-in end
Tension-only cable linkage
A uniform distributed load q acts over the middle 2a segment of beam AB of total length 4a. The equivalent resultant force P equals qa and acts at the segment’s center. What is the moment of P about point A, taking counterclockwise as positive?
qa·a counterclockwise
qa·2a counterclockwise
qa·2a clockwise
qa·a clockwise
A beam AB of length 4a carries a uniform load q from A to B. The resultant force is P=qa acting at distance 2a from A. What is the moment of the distributed load q about point A (counterclockwise positive)?
−4qa2
2qa^2
−2qa2
3qa^2
Beam ABC has segments AB=a and BC=2a. A uniform load q acts only over BC. Its resultant P=qa acts at the centroid of BC, a distance a from C toward B. What is the moment of P about point C, taking counterclockwise as positive?
2qa^2 counterclockwise
3qa^2 counterclockwise
qa^2 counterclockwise
−qa^2 clockwise
A prismatic beam AB of length 4a is supported at A and B, with a uniformly distributed load q acting downward on the right half segment BC of length 2a. Point C is at the right end. Determine the moment of the distributed load q about point C. Take clockwise as negative.
2qa^2
3qa^2
qa2
−qa2
A cantilever beam ABC fixed at A has total length 3a. A concentrated load P acts downward at point B located at distance a from A, and a uniformly distributed load q acts downward on the free end segment BC of length 2a. Determine the moment of the force P about point A. Take counterclockwise as positive.
−qa2
Cqa^2
2qa
qa
A prismatic beam ABC of length 3a is fixed at A and carries a uniformly distributed load q downward over the entire span BC of length 2a to the right of A. Determine the moment of the distributed load q about point A. Take clockwise as negative.
−4qa2
2qa^2
3qa^2
−3qa2
A straight bar ABC lies along the x-axis. Two vertical forces act in the Az direction: 5 kN upward at a point 2 m from A, and 10 kN downward at point C which is 3 m from A. Determine the reaction force at A in the Az direction (positive upward) assuming only vertical forces and supports along z.
−20 kN
15 kN
5 kN
−10 kN
A beam AB is supported and loaded as shown: a vertical force F acts downward at a point B a distance a from A. Determine the moment of force F with respect to point A. Take counterclockwise as positive.
Fa clockwise
Fa counterclockwise
F/a counterclockwise
F a^2 counterclockwise
For the beam shown, a force F = 500 N acts at point A with perpendicular distance 3 m to point B and 2 m to point S along the beam as indicated. Determine the moment of F about point A. Choose the correct value and sign convention shown.
-1500 (N·m)
1000 (N·m)
500 (N·m)
-1000 (N·m)
Beam ABC is loaded and supported as shown. Given F = 8 N, compute the moment of force F about point A.
-32 (N·m)
-8 (N·m)
-16 (N·m)
16 (N·m)
Beam ABC with member CD is loaded as shown. Let S be the internal force in member CD. What is the moment of force S about point A? Distances AB and AC are a (m).
-S a √2 / 2 (N·m)
S a (N·m)
-S a (N·m)
√2 S a (N·m)
Let S1 be the internal force in bar AB as shown. What is the moment of force S1 about point E?
S1·2a
S1·a
2 S1·a
S1·a/2
A force S acts along bar BD at 60° above the x-axis. Point C lies a distance a from point B along the x-axis. What is the moment of S about point C?
−S·sin(60°)·a
S·cos(60°)·a
S·a/2
S·√3·a
For the same bar-and-force system, what is the moment of the force S about point B? Assume the line of action of S passes through B.
0
S·a
−S·a
2S·a
A couple of magnitude M acts in the plane of the bar system with axis perpendicular to the plane. What is the moment of this couple about point B?
M
M·2a
−M·2a
0
At a pin joint A in three-dimensional space with coordinate axes Axyz, how many independent reaction force components can the support exert?
Three force components only
Two force components only
Three force and three moment components
One resultant force component only
Two force and one moment components
A prismatic beam ABC is loaded on segment AC by a linearly varying distributed load from 4 kN/m at A to 8 kN/m at C over length 3 m. Taking point A as reference, what is the net moment of this distributed load system about A? Use clockwise negative convention.
−12 kN·m
−18 kN·m
12 kN·m
24 kN·m
Two bars AB and CD are connected as shown. Let S1 and S2 be the internal axial forces in bars AB and CD, respectively. Considering equilibrium along axis x, what is the sum of the projections of all forces on x for the pin-jointed linkage BEK?
−S1 − S2
S1 + S2
0
S1 − S2
A shaft carries two forces F1=500 N and F2=300 N acting at radii r1=80 mm and r2=300 mm, respectively, as shown. What is the total moment of F1 and F2 about the z-axis? Use N·mm units and right-hand rule.
16000 N·mm
40000 N·mm
−40000 N·mm
−16000 N·mm
For three concurrent forces T1, T2, and Q acting at point A, the x-axis is to the right. Force T1 makes 45° above the negative x-axis, T2 acts upward along the positive y-axis, and Q acts downward along the negative y-axis. Which expression gives the sum of projections of these forces on the x-axis?
T2·√3/2 − T1·√2/2
T2·√2/2 + T1·√2/2
T1·√3/2 − T2·√2/2
−T1·√2/2 + 0 + 0
A rigid bar ABC is acted on by three forces T1 at A horizontal to the left, T2 at A vertical upward, and T3 at C making 30° above the positive x-axis to the right. What is the algebraic sum of the projections of T1, T2, T3 onto the x-axis? Take rightward components as positive.
−T1 − T3/2
T1 + T2 − T3/2
T1 − √3 T2/2
T1 − T2 − T3
When the reference point for the reduced system of forces is changed (i.e., you choose a new reduction pole), how does the principal vector of the force system change?
It does not change for any pole
It becomes zero for a suitable pole
It decreases by exactly one half
It reverses direction but keeps magnitude
Calling R the resultant vector of a force system reduced at point A. When changing the reduction point from A to B, the principal moment changes by an amount equal to:
M = m_AB (R̄)
M̄ = m_AB (R̄)
No change
M̄ = m_BA (R̄)
Let the principal vector and principal moment of the force system be R̄ and M̄. The general equilibrium conditions of a force system are:
R̄ = 0 & M̄ = 0
R̄ ≠ 0 & M̄ = 0
R̄ = 0 & M̄ ≠ 0
R̄ ≠ 0 & M̄ ≠ 0
For a rigid body in space under a general force system, how many equations can be written to determine the equilibrium conditions in the Oxyz coordinate system?
6
5
3
2
A rigid body under coplanar forces in the Oxy plane requires how many equations to determine equilibrium conditions?
3
4
2
1
A rigid body under a system of parallel forces in space requires how many equations to determine equilibrium conditions in Oxyz?
3
6
5
4
A rigid body under a concurrent force system in space requires how many equations to determine equilibrium conditions in Oxyz?
3
2
4
1
For a rigid body under a concurrent force system in the Oxy plane, how many independent equilibrium equations are required to determine balance?
Two independent equations are required
Three independent equations are required
One independent equation is required
Four independent equations are required
When shifting the force system’s reduced point from O to A, how does the scalar product between the principal resultant vector and the principal moment vector change?
It does not change in magnitude
It decreases by a factor of two
It increases by a factor of two
It becomes zero at any point
A force system is reduced to a wrench when the principal vectors R̄ and M̄ satisfy which condition?
R̄ ≠ 0 and M̄ ≠ 0
R̄ = 0 and M̄ = 0
R̄ = 0 and M̄ ≠ 0
R̄ ≠ 0 and M̄ = 0
A uniform beam AB of length 4a is subjected to a downward distributed load q over segment BC of length 2a from the midpoint, a point load P at location a from A, and support reactions Y (vertical at A) and N (vertical at C). Which Ay-direction equilibrium equation is correct?
Y − P − q·2a + N = 0
Y + P − q·2a + N = 0
Y − P + q·2a + N = 0
Y + P + q·2a + N = 0
For the same beam, taking moments about point A (counterclockwise positive), which moment equilibrium expression is correct if the distributed load over 2a has its resultant q·2a acting at its centroid located at distance 3a from A, and point load P acts at distance a from A, with reaction N at C at distance 4a from A?
N·4a − P·a − q·2a·3a = 0
N·3a − P·a − q·2a·2a = 0
N·4a − P·2a − q·2a·a = 0
N·2a − P·a − q·2a·3a = 0
For the beam ABC fixed at A with distributed load q over length 2a and a point force P at distance a from A, which equilibrium equation along the Ay direction is correct?
Y − P − q·2a = 0
Y − P − q·2a − M = 0
Y − P − q·2a·2a = 0
Y − P + q·2a = 0
For the same beam, which moment equilibrium equation about point A is correct if a clockwise positive sign convention is used?
M − P·a − q·2a·2a/2 = 0
M − P·a − q·2a·2a = 0
Y·a − P·a + M = 0
M − P·a + q·2a·2a/2 = 0
A beam AB is supported by reactions Y at A and N at C, with a uniform load q over 3a and a point force P at A. What is the correct moment equilibrium about point C?
P·2a − Y·3a − q·2a·a = 0
P·2a + Y·3a − q·2a·a = 0
N·3a + P·a + q·2a·2a = 0
N·3a + P·a + 3a·2a = 0
For a simply supported beam AB with reactions Y at A and N at B under a uniform load q over length 3a, what is the value of reaction N when the beam is in equilibrium?
N = (5/3)·q·a
N = (4/3)·q·a
N = (2/3)·q·a
N = (q·a)
N = (q·a)/2
For the beam AB with a uniform load q over length 4a and a point load N at the right end (as shown), the vertical reaction at support A is Y. What is the value of Y for equilibrium of AB?
Y = 4/3 qa
Y = 5/3 qa
Y = 1/3 qa
Y = qa
Consider a cantilever bar ABC fixed at A with distributed load q over segment BC of length 2a and a point load P = qa at B. Taking moments about point B, which equation expresses moment equilibrium?
M − Y·a − q·2a·a = 0
M − P·a + q·2a·2a = 0
M − P·a − q·2a·2a − Y·a = 0
M − P·a − q·2a·2a + Y·a = 0
For the cantilever ABC fixed at A with the same loading, determine the reaction moment M at A when segment AC is in equilibrium. Assume P = qa and distributed load q acts over length 2a.
M=5qa2
M = 5qa
M = 3qa
M=3qa2
A beam AB is supported with reaction components X, Y, T and subjected to an applied force F at an angle. Which equation represents force equilibrium along the x-direction (Ax)?
ΣFx = 0 gives X − F cosθ = 0
ΣFx = 0 gives X + F sinθ = 0
ΣFx = 0 gives X − F = 0
ΣFx = 0 gives X + T − F cosθ = 0
For the bar AB with reaction components X, Y, T and an applied force F as shown, what is the correct moment-equilibrium equation about point B?
F·2 − Y·5 = 0
F·2 − Y·5 − T·5 = 0
F·2 − Y·5 − F·2 = 0
F·2 + Y·5 = 0
For the same bar, which equation represents the moment equilibrium about point A?
T·sin(30°)·5 − F·3 = 0
T·cos(30°)·5 − F·3 = 0
T·cos(30°)·5 − F·3 + Y·5 = 0
T·sin(30°)·5 − F·3 + Y·5 = 0
Given the bar AB with reactions X, Y, T and force F = 500 N as shown, what is the value of the reaction Y when the system is in equilibrium?
Y = 200 (N)
Y = 500 (N)
Y = −500 (N)
Y = −200 (N)
A rigid bar AB is supported at A by pin reactions X and Y, and at C by a link BC carrying axial force T along BC at 30° above the horizontal. A vertical force F = 500 N acts downward at midspan between A and B. Distances: A–midpoint = 2 m, midpoint–B = 2 m, and C is at the right end. When the bar is in equilibrium, what is the value of the reaction T in link BC?
T = 600 N
T = 300 N
T = 300√3 N
T = −300/√3 N
For the same bar AB with reactions X, Y at A and a support force S at D acting along a 45° direction to the negative x-axis, a vertical force F = 500 N acts downward at the midpoint. Write the correct force equilibrium equation along the x-axis.
X − S·cos(45°) = 0
X − S = 0
X − S·cos(45°) − F = 0
X − S·cos(45°) + F = 0
Consider bar AB with pin reactions X, Y at A and an inclined support force S at D, with a downward force F = 500 N acting at a point 2 m from A and 2 m from B. Taking moments about point C located 2 m from B toward A, which equilibrium moment equation is correct?
−Y·2 − F·2 = 0
Y·2 − F·2 = 0
Y·2 − F·2 + X·2 = 0
Y·2 − F·2 − X·2 = 0
For the configuration with reactions X, Y at A, an inclined support S at D, and a downward force F = 500 N at the midpoint, what is the vertical reaction Y at support A when the bar is in equilibrium?
Y = 0 N
Y = 250 N
Y = 500 N
Y = −250 N
Bar AB is supported at A and B with reaction components X and Y at B and A, and an inclined reaction S at point near A as shown. A horizontal force F = 500 N is applied at B. Distances are: A to load point C is 2 m, C to B is 2 m, and the vertical height of A above the lower reference is 2 m. When the system is in equilibrium, the magnitude of the reaction S is:
S = 1000√3 (N)
S = −1000√3 (N)
S = −500√2 (N)
S = 500√2 (N)
For the same bar AB with reactions X, Y, S and applied force F = 500 N at B, write the scalar moment equilibrium equation about point A using the forces shown. Choose the correct expression:
−S·sin(45°)·2 − F·4 = 0
S·sin(45°)·2 − F·4 = 0
−S·sin(45°)·2 + F·4 = 0
S·√3/2 + F·4 = 0
Member BDEK has support reactions S1, S2, and N and external forces as shown. What is the correct equilibrium equation for the sum of forces in the vertical direction, ΣEy = 0?
S1 + S2 − N = 0
S1 − S2 − N = 0
S1 + S2 + N = 0
S1 − S2 + N = 0
For the vertical member BDEK, the support reactions are S1 at B, S2 at D, and a normal reaction N at the guide near E. A horizontal force F acts at K on the right, at height a above E, and the link distances are shown with a square layout of side a. Writing the moment equilibrium about point E (taking counterclockwise positive), which expression is correct?
S1·2a + S2·a − F·a = 0
S1·2a − S2·a + F·a = 0
S1·a + S2·a − F·2a = 0
S1·2a + S2·a + F·a = 0
Considering the same frame, write the moment equilibrium about point D for the forces S1 at B, S2 at D, the normal reaction N near E, and a horizontal force F applied at K at a vertical distance a above E. Which relation satisfies ΣM_D = 0 with counterclockwise positive?
S1·a − F·a = 0
S1·2a + S2·a + F·a = 0
S1·a + F·a = 0
S1·2a − S2·a − N·a = 0
In the planar structure BDEK, the external force is F = 100 N acting horizontally to the right at point D. The bar is supported by a pin at E, producing a reaction with horizontal component N only. When the system is in equilibrium, what is the value of reaction N at E?
N = 100 N
N = 0 N
N = 100.4 N
N = −100.4 N
Consider a rigid bar AB of length 2 m connected to the frame BDEK by pins at A and B. A horizontal force F = 100 N acts at D to the right. Reaction forces at A are S1 (horizontal) and S2 (vertical). When the system is in equilibrium, what is the value of S1 at A?
S1 = 100 N
S1 = 0 N
S1 = 50 N
S1 = 200 N
For the same system as above, with F = 100 N to the right and supports providing components S1 and S2 at A, determine the vertical reaction S2 at A when the system is in equilibrium.
S2 = −100 N
S2 = −1000 N
S2 = 50 N
S2 = 200 N
For the beam ABC with unknown reaction components S, Xc, Yc and a force S inclined at 60° to the positive x-axis at point D as shown, the equilibrium equation for the projection of forces on axis Bx is:
S·cos(60°) − Xc = 0
S·sin(60°) − Xc = 0
S·sin(60°) − Xc + Y = 0
S·cos(60°) − Xc + Y = 0
For the same beam, the equilibrium equation for the projection of forces on axis By is:
S·sin(60°) + Yc − q·2a = 0
S·sin(60°) − Yc − q·2a = 0
S·cos(60°) + Yc − q·2a = 0
S + Yc − q = 0
Taking moments about point C for all external forces on the beam, the correct scalar moment equilibrium equation is:
M + q·2a·2a − S·sin(60°)·a = 0
M + q·2a·2a − S·cos(60°)·a = 0
M + q·2a·2a + S·cos(60°)·a = 0
M + q·2a + S·sin(60°)·a = 0
Given M = 5 kN·m, q = 2 kN/m, and a = 2 m for the beam shown, the magnitude of the inclined reaction S at member BD that satisfies equilibrium is:
S = 37/√3 (N)
S = 37/(√3) (N)
S = 37√3 (N)
S = 37 (N)
With the same data (M = 5 kN·m, q = 2 kN/m, a = 2 m), the horizontal reaction at C is:
Xc = 37/(2√3) (N)
Xc = 37/√3 (N)
Xc = 37√3 (N)
Xc = 37 (N)
With the same data, the vertical reaction at C is:
Yc = −21/2 (N)
Yc = −21/4 (N)
Yc = −21/√3 (N)
Yc = 21/√3 (N)
For beam ACB with vertical reaction Y at A and vertical reaction N at B, carrying a uniformly distributed load q1 = 4 kN/m over the 3 m span AC, which correct force projection equilibrium along Ay is satisfied? Assume upward positive and the load acts downward over 3 m.
Y + N − q1·3 = 0
Y − N − q1·3 = 0
Y + N + q1·3 = 0
Y − N + q1·3 = 0
For the same beam, taking moments about point A with counterclockwise positive, which moment equilibrium equation is correct? The resultant of the distributed load is q1·3 acting at 1.5 m from A.
N·6 − q1·3·1.5 = 0
N·6 − q1·3·3 = 0
N·6 + q1·3·1.5 = 0
N·3 − q1·3·1.5 = 0
Using equilibrium, what is the reaction force N at support B? Take the span AB = 6 m and the distributed load q1 = 4 kN/m over the first 3 m from A.
N = 3 kN
N = 6 kN
N = 9 kN
N = 12 kN
What is the reaction force Y at support A for the same loading? Use vertical force and moment equilibrium with q1 = 4 kN/m on 3 m and span AB = 6 m.
Y = 3 kN
Y = 6 kN
Y = 9 kN
Y = 12 kN
For the force system at point A with forces T1 at 45° above +x, T2 at 30° above −x, and a downward force Q, the equilibrium equation along Ax is:
T2·cos(30°) − T1·sin(45°) = 0
T2·sin(30°) + T1·sin(45°) = 0
T2·sin(30°) − T1·sin(45°) = 0
T2·sin(45°) + T1·sin(30°) = 0
For the same system, the equilibrium equation along Ay is:
T1·sin(45°) + T2·sin(30°) − Q = 0
T1·cos(45°) + T2·cos(30°) − Q = 0
T1·cos(45°) − T2·cos(30°) − Q = 0
T1·cos(45°) + T2·cos(30°) + Q = 0
Given Q = 300 N acting downward and the same directions for T1 (45°) and T2 (30°), the equilibrium yields the value of T1 as:
T1 = 300 / (√3 + √2/2) (N)
T1 = 600 / √3 (N)
T1 = 300 / √3 (N)
T1 = 300 / √2 (N)
For a beam ABC fixed at A, with forces T1 vertical at AB, T2 inclined left at 45° at A, and T3 along BC at 30° above +x, the Ax equilibrium equation using support reaction X at A is:
X − T1 − T3·sin(30°) = 0
X − T1 − T2·cos(30°) = 0
X − T1 + T3·cos(30°) = 0
X − T1 + T3·sin(30°) = 0
For the same beam, the Ay equilibrium equation with vertical reaction Y at A is:
Y − T2 − T3·cos(30°) = 0
Y − T2 − T3·sin(30°) = 0
Y − T2 + T3·cos(30°) = 0
Y + T2 − T3·cos(30°) = 0
Which statement correctly describes the role of reaction moment M at a fixed support in planar equilibrium for the beam ABC?
M balances net moment about A from external forces
M balances only horizontal components of all forces
M replaces shear force and axial force at the section
M eliminates the need for X and Y at the support
In 3D space, a point M moves with x = x(t), y = y(t), z = z(t). Which expression gives the magnitude of the total velocity v_M?
vM=x′2+y′2+z′2
vM=x2+y2+z2
v_M = x' + y' + z'
vM=x2+y2+z2
In 3D space, a point M moves with x = x(t), y = y(t), z = z(t). Which expression gives the magnitude of the total acceleration a_M?
a_M = x′′2+y′′2+z′′2
aM=x′2+y′2+z′2
a_M = x'' + y'' + z''
aM=x′2+y′2+z′2
Which statement is correct about differentiating the position vector r(t) of a moving point?
Velocity equals the first time derivative of the position vector
Acceleration equals the first time derivative of the position vector
Position equals the first time derivative of the position vector
Velocity equals the second time derivative of the position vector
Given x = x(t), y = y(t), z = z(t), what is this set of equations called in Cartesian coordinates?
Equations of motion of the point in Cartesian coordinates
Velocity components of the point in Cartesian coordinates
Acceleration components of the point in Cartesian coordinates
Trajectory equation of the point in Cartesian coordinates
If you eliminate t from x = x(t), y = y(t), z = z(t) to relate x, y, z, what do you obtain?
The trajectory equation of the point
The position vector of the point
The coordinate frame definition
The differential equation of motion
In natural coordinates, the total acceleration of a point M is given by which expression, where v is speed, ρ is radius of curvature, t̂ is the unit tangent, and n̂ is the unit normal?
a = v̇ t̂ + v²/ρ n̂
a = v̇ t̂
a = v²/ρ n̂
a = v̇ t̂ + v²/ρ ī
For motion in natural coordinates, the correct expression for the normal (centripetal) component of acceleration is:
a_n = v²/ρ n̂
a_n = v̇ n̂
a_n = v̇ t̂
a_n = v̇ t̂̄
Which statement is correct about the normal component of acceleration in natural coordinates?
Its direction is from the point toward the center of curvature
It directly indicates the magnitude of the velocity vector
It describes the entire motion of the particle
It models uniformly accelerated rectilinear motion
A point moves as x = 3 cos(4t), y = 3 sin(4t), z = 2t (meters, seconds). What is the trajectory of the point?
A circular helix of pitch 2 meters
A circle of radius 3 centered at O
A parabola in the x–y plane
A line given by y = 3x + 4
A point moves as x = 4 cos(2t), y = 3 sin(2t), z = 0 (meters, seconds). What is the trajectory of the point?
An ellipse with semi-axes 4 m and 3 m
A circular helix of pitch 3 meters
A circle of radius 4 meters
A line along the x-axis
A point M moves on a circle of radius 6 cm with position given by s=6t+9t2+12t3 (cm), where s is arc length along the circle and t in seconds. What is the speed of M at t = 2 s?
180 cm/s
18 cm/s
170 cm/s
17 cm/s
A point moves in the plane with parametric equations x = 75 cos(4t^2), y = 75 sin(4t^2), with t in seconds and x,y in cm. What is the correct expression for the instantaneous speed v(t)?
v = 600t cm/s
v = 600 cm/s
v = 75 t^2 cm/s
v = 75 t cm/s
Given the motion x=75cos(4t2) , y=75sin(4t2) (cm), which statement best describes the trajectory of the point?
A circle centered at (0,0) of radius 75 cm
A helix with pitch equal to 4 cm
A straight line given by y = 75 x + 4
A straight line given by y = 4 x + 75
A point M moves on a circle of radius 10 cm with arc-length function s=4t3+2t (cm). What is the speed of M at t = 5 s?
302 cm/s
300 cm/s
30.2 cm/s
30 cm/s
