A crane consists of a tower and (massless) crossbeam, as shown. The crossbeam pivots about a hinge at point P. The counterweight M (10³ kg) is located d = 10 m to the left of the hinge, and the load mass m is located x = 25 m to the right of the hinge. A cable is attached to the right end of the crossbeam, l = 30 m from the hinge at an angle of 35°. The maximum tension that the cable can withstand before breaking is 2 x 10⁴ N. What is the maximum mass m that can be supported?

Solution:

This is clearly a statics problem, with the crossbeam as the object in question. The way to attack these problems is always the same - draw a force diagram, choose a coordinate system and a point to take torques about, and set the sum of the forces and the sum of the torques both equal to zero. Then it should be just a matter of algebra.

The forces on the crossbeam include the weights Mg and mg, the tension in the cable T , and the hinge force F. We don't know the direction of the hinge force, so we can consider it as two separate forces F_x and F_y in the horizontal and vertical directions respectively. We will choose the tension as the maximum value 2 x 10⁴ N, since we want to find the maximum m that can be lifted.

If we are crafty, and take the torque about point P, then the torque equation will not involve the hinge force (since it acts at point P). In this case the torque equation involves only one unknown, m, so we don't even need the force equations, and the algebra is trivial! We have

= Mgd + Tsinl -mgx = 0

Here, the lever arms for the weights Mg and mg are clearly just d and x respectively, and I have chosen counterclockwise as the direction for positive rotations. The torque caused by the tension is Tsinl, which can be obtained in three equivalent ways:

1. Consider r x F, where r = l, and F = T. The magnitude of this torque is then Tlsin, where is the angle between T and l, which is 180° - 35°, but the sine of 180°- is the same as the sine of .
   or
2. Break T into horizontal (-Tcos35°) and vertical (Tsin35°) components. The line of action of the horizontal component goes through P, so it causes no torque; the vertical component has a lever arm of l, so the total torque is just l(Tsin35°).
   or
3. Drop a perpendicular from P to the line of action of the tension (which extends in the direction of the cable). The length of this perpendicular distance is, by defintion, the moment arm; from the diagram its length is lsin35°, so the torque is T(lsin35°).

Solving for m gives

m = [Mgd + Tsinl]/gx

= [(10³ kg)(9.8 m/s²)(10 m) + (30 m)(2 x 10⁴)sin35°]/[(9.8 m/s²)(25 m)]

m = 1805 kg

The same result could have been obtained if we had chosen to take the torque about any other point than P, but then the torque equation would involve the hinge force components, and one would need to use the x and y force equations to get three equations in three unknowns, which can be solved for m (and F_x and F_y) - just a bit more algebra.

Note:

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