Question

A crane has an arm length of 8.1 m inclined at 30o with the vertical. It carries a container of mass of 2.6 kg suspended from the top end of the arm. Find the magnitude of the torque produced by the gravitational force on the container about the point where the arm is fixed to the crane. (Neglect the weight of the arm)

Answer 1

Given:

arm length of crane r =  8.1 m

inclination angle θ = 30°

mass of container m = 2.6 Kg

To find:

the magnitude of the torque τ =?

Step-to-step-explanation:

• A crane has an arm length of 8.1 m which is inclined at 30° with the vertical.
• The mass of 2.6 kg is suspended from the top end of the arm.
• Let the force be F at the point of suspension is due to the weight of the hanging mass.
• This suspension force is given by:

        F = mg

           =  2.6 × 10

           = 26 N

• From the figure, this suspension force is perpendicular to the arm length of a crane.

        τ  = r ⊥ F

           = rf cosθ

           = 8.1  × 36  × cos 30°

       τ  = 182.3849 Nm

       τ  ≈ 182 Nm

• Hence the magnitude of the torque produced by the gravitational force on the container is 182 Nm.