A metal bar is in the xy-plane with one end of the bar at the origin. A force F

Question

A metal bar is in the xy-plane with one end of the bar at the origin. A force F rightarrow = (6.96 N)i^ + (-3.17 N)j^ is applied to the bar at the point x = 3.99 m, y = 3.84 m. Part A What is the position vector? for the point where the force is applied? Enter the x and y components of the radius vector separated by a comma. My Answers Give Up Part B What are the magnitude of the torque with respect to the origin produced by f? Express your answer with the appropriate units. PartC What are direction of the torque with respect to the origin produced by frightarrow?

Explanation / Answer

Here,

F = 6.96 i - 3.17 j N

part A)

for the vector r

r = 3.99 i + 3.84 j

(rx , ry) = (3.99 , 3.84) m

part B)

for the torque applied

Torque = r X F

Torque = (3.99 i + 3.84 j ) X (6.96 i - 3.17 j)

Torque = (3.99 * 6.96 + 3.84 * 3.17) k N.m

Torque = 39.94 k N.m

the magnitude of torque acting is 39.94 N.m

part c)

the direction of torque is +z direction

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