91影视

The counterweight placed on the crane is\(M = 9500\;{\rm{kg}}\).

The mass lifted by the crane is\(m = 2800\;{\rm{kg}}\).

The horizontal length of the crane is\(d = 7.7\;{\rm{m}}\).

The distance at which the counterweight lies is\(d' = 3.4\;{\rm{m}}\).

In this problem, the crane is in equilibrium. Hence, its net torque is zero. For calculating the maximum load, use the sum of all torques about the pivot point.

The following is the free-body diagram of the crane.

The relation for net torque is shown below.

\(\begin{array}{c}\sum \tau = 0\\\left( {Mg \times x} \right) - \left( {mg \times d} \right) = 0\end{array}\)

Here,\(g\)is the gravitational acceleration, and\(x\)is the required distance.

Put the values in the above relation.

\(\begin{array}{c}\left( {\left( {9500\;{\rm{kg}}} \right)\left( {9.8\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}} \right) \times \left( x \right)} \right) - \left( {\left( {2800\;{\rm{kg}}} \right)\left( {9.8\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}} \right) \times \left( {7.7\;{\rm{m}}} \right)} \right) = 0\\x = 2.26\;{\rm{m}}\end{array}\)

Thus, \(x = 2.26\;{\rm{m}}\) is the required distance for the placement of the counterweight.

The relation for the maximum load is

\(\begin{array}{c}\sum \tau = 0\\\left( {Mg \times d'} \right) - \left( {m'g \times d} \right) = 0.\end{array}\)

Here,\(m'\)is the mass lifted by the crane.

Put the values in the above relation.

\(\begin{array}{c}\left( {\left( {9500\;{\rm{kg}}} \right)\left( {9.8\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}} \right) \times \left( {3.4\;{\rm{m}}} \right)} \right) - \left( {\left( {m'} \right)\left( {9.8\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}} \right) \times \left( {7.7\;{\rm{m}}} \right)} \right) = 0\\m' = 4194.8\;{\rm{kg}}\end{array}\)

Thus, \(m' = 4194.8\;{\rm{kg}}\) is the maximum load lifted by the crane.