Question

A large crane consists of a 20 m, 3,000 kg arm that extends horizontally on top of a vertical tower. The arm extends 15 m towards the lifting end and 5 m towards the counterweight. If the crane is to lift a 5,000 kg load, what must the weight of the counterweight be in order to maintain static equilibrium

Answer 1

`m=18000kg`

Step-by-step explanation:

From the question we are told that:

Crane Length
`l=20m`

Crane Mass
`m_a=3000kg`

Arm extension at lifting end
`l_l=15m`

Arm extension at counter weight end
`l_c=5m`

Load
`M_l=5000kg`

Generally the equation for Torque Balance is mathematically given by

`T_1 *l_c-(m_a*g) *l_c-(T_2)*l_l=0`

`mg*5 *-(3000*9.8) *5-(5000*9.8)*15=0`

`m=18000kg`