Question

A 3.0-m rod is pivoted about its left end. A force of 6.0 N is applied perpendicular to the rod at a distance of 1.2 m from the pivot causing a ccw torque, and a force of 5.2 N is applied at the end of the rod 3.0 m from the piVOT?

Answer 1

The question involves calculating the net torque on a rod pivoted at one end due to two different forces. The torque is the product of force, distance from the pivot, and the sine of the angle between them.

Step-by-step explanation:

The question asks about the torque created by forces acting on a rod that is pivoted at one end. Torque is a measure of the tendency of a force to rotate an object about an axis. In this case, one force is applied at a distance of 1.2 m from the pivot, and another force is applied at the end of the 3.0-m rod. To find the net torque, you would subtract the torque due to the 5.2 N force from the torque due to the 6.0 N force since they cause rotation in opposite directions. The torque (τ) is calculated using the equation τ = r × F × sin(θ), where r is the distance from the pivot, F is the force applied, and θ is the angle between the force and the lever arm (which is 90° or π/2 radians in this case, making sin(θ) = 1).

Answer 2

-0.6Nm

Step-by-step explanation:

Torque is defined as the perpendicular component of force \times distance from pivot point.

#
`T_n_e_t=\sum{T_i}` where
`T_i` is the torques from different applied forces.

`T_1=6N*1.2m=7.2Nm` (+ve, counterclockwise)

`T_2=-5.2N* sin30\textdegree* 3m\\=-7.8Nm`(-ve, clockwise)

`T_n_e_t=T_1+T_2\\=(7.2-7.8)Nm\\=-0.6Nm`

Hence, the total torque is -0.6Nm