A crane has an arm length of 20 m inclined at 30° with the vertical. It carries a container of mass of 2 ton suspended from the top end of the arm. Find the torque produced by the gravitational force on the container about the point where the arm is fixed to the crane. [Given: 1 ton = 1000 kg, neglect the weight of the arm. `g-10 m s^(-2)]`

Explanation

The force at the point of suspension is due to the weight of the hanging mass
`F=mg=2xx100xx10=20000 N`
The are length r=20 m
We can solve this problem by three different methods.
Method-1: The angle () between the arm length() and the force (F) is `theta=150^(@)` The torque (t) about the fixed point of the arm is,
`tau=r F sin theta`
`tau=20xx2000xx sin (150^(@))`
`=400000xx sin (90^(@)+60^(@)) " " [ "here" sin (90^(@)+theta)= cos theta]`
`=400000xx(1)/(2)[ cos 60^(@)=(1)/(2)]=200000 Nm`
`tau=2xx10^(5) Nm`
Method-II: Let us take the force and perpendicular distance from the point where the arm is fixed to the crane
`tau=(r bot)F`
`tau= r cos phii mg`
`tau=20xx cos 60xx20000`
`=20xx(1)/(2)xx20000=200000 Nm`
`tau = 2xx10^(4)Nm`

Method-III: Let us take the distance from the fixed point and perpendicular force
`tau=(F bot)`
`tau= r mg cos phi`
`ta=20xx20000xx cos 60^(@)` `tau= 2x10^(5) Nm`
All the three methods, give the same answer