Answer to Question #86862 in Mechanics | Relativity for Aiswarya achu

Question

A 1000 kg truck is parked on a bridge 100 m long at a distance of 25 m from one end of
the bridge. The mass of the bridge is 200 kg m−1
. Calculate the reaction forces at the
supports at the two ends of the bridge. (Take g = 10 ms−2
)

Answer 1

$$\sum M_A=0\\ R_BL=W_{\rm{truck}}d+W_{\rm{bridge}}L/2$$

$$R_B=\frac{W_{\rm{truck}}d+W_{\rm{bridge}}L/2}{L}=\frac{1000\times 10\times 25+200\times 10\times 50}{100}$$

$$R_B=3500\:\rm{N}$$

$$\sum M_B=0\\ R_AL=W_{\rm{truck}}(L-d)+W_{\rm{bridge}}L/2$$

$$R_A=\frac{W_{\rm{truck}}(L-d)+W_{\rm{bridge}}L/2}{L}=\frac{1000\times 10\times 75+200\times 10\times 50}{100}$$

$$R_A=8500\:\rm{N}$$