# Statics - Forces on a crane

Problem Statement:

The figure below shows a crane (for simplicity we will neglect its mass) whose arms have lengths d and 2d respectively. A mass m is fixed to the left and another mass m can move horizontally along the arms of the crane. Another mass M is attached to the end of the right arm. The system is in static equilibrium. Determine:

1. The distance x with respect to the origin O to which the moving mass m must be situated if the mass M = m/4.
2. The value of M if the moving mass m is at the origin of coordinates.
3. The maximum value of M that the crane can hold. Knowledge is free, but servers are not. Please consider supporting us by disabling your ad blocker on YouPhysics. Thanks!

Solution:

To solve this problem we will use the condition of static equilibrium so that a system does not rotate (equation of rotation): We first draw the forces acting on the system. The only relevant forces for solving the problem are the weights of the different masses, we do not need to take into account the tension of the cables or the reactions because they are internal forces. For simplicity, we consider masses as point particles.

In problem 1 we saw that in order to calculate the moments of the masses we can neglect the ropes from which they hang, so in the following figure we will represent them attached to the crane. The equation of rotation for this system is: We will now calculate the torques that appear in the previous equation.

Weight P1: The torque of P1 is: We have represented both vectors in the following figure: Using the right-hand rule, we determine that the direction of the cross product is perpendicular to the screen and outward (k).

The torque magnitude is given by: Weight P2: The torque of P2 is: We have represented both vectors in the following figure: Using the right-hand rule, we determine that the direction of the cross product is perpendicular to the screen and inward (-k).

The torque magnitude is given by: Weight P: The torque of P is: We have represented both vectors in the following figure: Using the right-hand rule, we determine that the direction of the cross product is perpendicular to the screen and inward (-k).

The torque magnitude is given by: Knowledge is free, but servers are not. Please consider supporting us by disabling your ad blocker on YouPhysics. Thanks!

In the following figure we have represented the torques of the three weights (they are not scaled) in three dimensions: This representation is useful for determining the signs of the projections of each vectors of the equation of rotation on the z axis. It is not necessary to project on the x and y axes because the torques have no components on these two axes.

The projection of the equation of the rotation on the z axis is then: And after replacing with the previously calculated magnitudes we get: This equation (1) will allow us to solve all the sections of the problem.

If M = m/4, we can determine the value of x by substituting in equation (1): If x = 0, we can determine the value of M using the same equation: The crane will hold the maximum value of M when the moving mass is at x = -d. Substituting in equation (1) we get: The post Statics - Forces on a crane appeared first on YouPhysics