Chapter 9: Q3P (page 230)
(I) A tower crane (Fig. 9–48a) must always be carefully balanced so that there is no net torque tending to tip it. A particular crane at a building site is about to lift a 2800-kg air-conditioning unit. The crane’s dimensions are shown in Fig. 9–48b. (a) Where must the crane’s 9500-kg counterweight be placed when the load is lifted from the ground? (The counterweight is usually moved automatically via sensors and motors to precisely compensate for the load.) (b) Determine the maximum load that can be lifted with this counterweight when it is placed at its full extent. Ignore the mass of the beam.
\(4194.8\;{\rm{kg}}\)
Short Answer
The location of crane’s counterweight is (a) \(2.26\;{\rm{m}}\), and the maximum load that can be lifted is (b) .
Step by step solution
Given Data
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\(d' = 3.4\;{\rm{m}}\).
Evaluate the net torque and maximum load
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.
Draw the free-body diagram and calculate the position of the counterweight
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.
Calculate the maximum load lifted with this 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.
Over 30 million students worldwide already upgrade their learning with Vaia!
