QUIZ REVIEW Components of Force/Ramps

What is the horizontal (x) component of a 60 N force at 30 degrees above surface?

• 30 N

• 52 N

• 42 N

• 18 N

Answer explanation

Cosine component

cos 30 * 60N = 51.96

Tags

NGSS.HS-PS2-1

What is the normal force for a 10 kg sled on a ramp at 30 degrees above surface?

(Hint: need to find weight first, then normal is equivalent to one of the components)

• 85 N

• 100 N

• 120 N

• 65 N

Answer explanation

Weight =10*9.8=98N

Normal is opposite the Perpendicular component

cos30 * 98 = 84.87

What is the component responsible for pulling a 20 kg suitcase down a ramp at 30 degrees above surface?

• 10 N

• 57 N

• 98 N

• 196 N

Answer explanation

Need weight first. W=20*9.8=196N

parallel component

sin30 * 196 N=98 N

Which of the following is the best expression for the Normal Force?

• F_N = mg cos(ϴ)

• F_N = mg sin(ϴ)

• F_N = mg sin(180-ϴ)

• F_N = mg cos(90-ϴ)

Tags

NGSS.HS-PS2-1

A late traveler rushes to catch a plane, pulling a suitcase with a force directed 30.0° above the horizontal. If the horizontal component of the force on the suitcase is 60.6 N, what is the force exerted on the handle?

(Hint: a new twist on our regular component problems, they give you component, can you find hypotenuse?)

• 96

• 53

• 70.

• 65

Answer explanation

cos30=60.6/hyp

hyp = 60.6/cos30

hyp or force on handle = 69.97 N

What is the x component of the vector given in the diagram?

• 77N
• 92N
• 120N
• 60N

Tags

CCSS.HSN.VM.A.2

What is the y component of the vector given in the diagram?

• 77N
• 92N
• 120N
• 60N

Tags

CCSS.HSN.VM.A.2

How would you find the magnitude of a resultant vector that has an x component of 8 N and a y component of 6 N?

• 
  

  $\sqrt{8^2+6^2}$

• $\sqrt{8^2-6^2}$

• $\tan^{-1}\left(\frac{8}{6}\right)$
  

• $\tan^{-1}\left(\frac{6}{8}\right)$
  

How would you find the angle of a resultant vector from rope forces acting north at 8 N and then east at 6 N?

• 
  

  $\sqrt{8^2+6^2}$

• $\sqrt{8^2-6^2}$

• $\tan^{-1}\left(\frac{8}{6}\right)$
  

• $\tan^{-1}\left(\frac{6}{8}\right)$
  

Answer explanation

angles go from the start.  so north, then east would yield a start point at tail of the north vector...Opposite is 6 and adjacent is 8. so  $\tan^{-1}\left(\frac{6}{8}\right)$  

What is the x component of the vector?

• 14cos(30)

• 14sin(30)

• 14tan(30)

• 14tan^-1(30)