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

Answer:

$m=18000kg$

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

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

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

$m=18000kg$