Answer to Question #101919 in Mechanics | Relativity for jude

Question

A 15m wide horizontal platform is suspended by ropes at each end, each capsule of withstanding tension of 112KN. The uniform concrete platform has mass 1500kg.
a) Calculate the maximum load that can be added to the centre of the platform
b) Calculate the maximum load that can be added to the extreme ends of the platform
c) Explain why these two calculations produce different values
d) Suggest a value for the maximum safe recommended load for the platform and justify

Answer 1

a) Place the maximum load at the center. To stay horizontal and not flip over, we need 4 ropes at every corner of the platform. According to Newton's second law, if everything is at equilibrium, nothing will move. The ropes will break (and the concrete platform will start moving) if the load exceeds the maximum tension, so, write the equilibrium condition:

$$4T=(M_P+M_L)g,\\ \space\\ M_L=\frac{4T}{g}-M_P=\frac{4\cdot112000}{9.8}-1500=44214\text{ kg}.$$

b) If we add equal loads to every corner of the platform, the sum of these 4 weights will be the same as $M_L$ in part a). However, if we put the load on just one end, the answer will be different. Assume that the axis of rotation comes through the other two ropes, hence:

$$-2TL+LM_L'g+\frac{M_PgL}{2}=0,\\ \space\\ M_L'=\frac{2T}{g}-\frac{M_P}{2}=22107\text{ kg}.$$

c) Because in the first case the load is spread across all 4 ropes. In the second situation, the load is "hanged" to only two ropes.

d) Since, as we saw, the previous result is the lowest, we can say that it is the maximum safe load. But if we put the load on just one corner, the situation will change. If the platform is a square, and if the pivot point is at the opposite corner:

$$-TL\sqrt{2}+M_L'gL\sqrt{2}+\frac{M_PgL\sqrt{2}}{2}=0,\\ \space\\ M_L'=\frac{T}{g}-\frac{M_P}{2}=10679\text{ kg}.$$

This is the maximum allowable and safe load.