Calculus III: The Cross Product (Level 9 of 9)

This video concludes the cross product series by exploring torque through various examples. It covers calculating torque using both component and geometric definitions, with examples involving vertical forces on levers, forces on rigid bodies, and determining maximum force on a wrench. The video emphasizes the relationship between angles, force vectors, and position vectors in torque calculations.

What is the z-component of the force vector when a vertical force of 50 pounds is applied downward?

• 100 pounds

• 0 pounds

• -50 pounds

• 50 pounds

In the first example, what is the y-component of the position vector when the angle is 60 degrees?

• 0

• Square root of 3

• 1

• 2

How does the torque vector point when the angle between the lever and the positive y-axis is pi/2?

• Into the page

• Negative x-direction

• Out of the page

• Positive x-direction

In the second example, what is the angle between the position vector and the force vector?

• 45 degrees

• 60 degrees

• 105 degrees

• 90 degrees

What is the magnitude of the torque in the second example?

• 36 ft-lb

• 197 ft-lb

• 100 ft-lb

• 50 ft-lb

In the final example, what is the maximum force that can be applied to the wrench?

• 50 pounds

• 106.4 pounds

• 150 pounds

• 200 pounds

Which angle is irrelevant to the calculation of torque in the final example?

• 50 degrees

• 90 degrees

• 70 degrees

• 180 degrees