In the figure given, the system is in equilibrium. What is the maximum value that W can have if the friction force on the 40N block can not exceed 12.0N?

Options

A. 3.45N
B. 6.92N
C. 10.35N
D. 12.32N

Answer

B

Explanation

The free-body diagram of the block is as shown in adjacent figure.
`because (f_(s))_("max")=12.0N` (Given)
`therefore(T_(1))_("max")=12.0N`
The free-body diagram of knot is as shown in figure.
`T_(2) sin 30^(@)=W` ....(i)
`T_(2) cos 30^(@)=T_(1)` ...(ii) Divide (i) by (ii), we get
`tan 30^(@)=(W)/(T_(1))`
or `W=T_(1) tan 30^(@)=0.577T1`
`therefore W_("max")=0.577(T_(1))_("max")`
`=0.577xx12.0N=6.92N`