i want part B to be solved A metal bar is in the xy-plane with one end of the ba

Question

i want part B to be solved

A metal bar is in the xy-plane with one end of the bar at the origin. A force F = (7.41 N)I + (-3.50 N)j is applied to the bar at the point x = 3.23 m, y = 3.37 m. What is the position vector r for the point where the force is applied? Enter the x and y components of the radius vector separated by a comma. What are the magnitude of the torque with respect to the origin produced by F? Express your answer with the appropriate units. What are direction of the torque with respect to the origin produced by F?

Explanation / Answer

Torque = rxF

rxF = (3.23 i + 3.37 j)x(7.41 i - 3.5 j)

Now we know that

ixi = 0

jxj = 0

ixj = k

jxi = -k

Using these

rxF = 3.32*7.41*(ixi) + 3.23*(-3.5)*(ixj) + 3.37*7.41*(jxi) + 3.37*(-3.5)*(jxj)

rxF = 0 - 11.3050 k - 24.9717 k

rxF = - 36.2767 k

Magnitude will be = 36.28 N-m

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