1
Question
A board of mass \(m=11~kg\) is placed on the floor and a man of mass \(m=70~kg\) is standing on the board as shown (Case\(−I\)). The coefficient of friction between the board and the floor is \(\mu=0.25\). The maximum force that the man can exert on the rope so that the board does not slip on the floor is:
Open in App
Solution
For man \(+\) board system,
\(f=2T\)
\(T+N=(M+m)g\)
\(N=(M+m)g-T\)
If board is not sliding
\(f\leq \mu N\)
\(2T\leq \mu[(M+m)g-T)]\)
\([2+\mu]T\leq \mu(M+m)g\)
\(T\leq \dfrac{\mu(M+m)g}{(2+\mu)}\)
Hence,
\(T_{max}=\dfrac{\mu(M+m)g}{2+\mu}\)
After substituting the values:
\(T_{max}=\dfrac{0.25(70+11)}{(2+0.25)}\times10\)
\(=\dfrac{0.25\times81\times10}{2.25}=90~N\)
For man \(+\) board system,
\(f=2T\)
\(T+N=(M+m)g\)
\(N=(M+m)g-T\)
If board is not sliding
\(f\leq \mu N\)
\(2T\leq \mu[(M+m)g-T)]\)
\([2+\mu]T\leq \mu(M+m)g\)
\(T\leq \dfrac{\mu(M+m)g}{(2+\mu)}\)
Hence,
\(T_{max}=\dfrac{\mu(M+m)g}{2+\mu}\)
After substituting the values:
\(T_{max}=\dfrac{0.25(70+11)}{(2+0.25)}\times10\)
\(=\dfrac{0.25\times81\times10}{2.25}=90~N\)
Suggest Corrections
1
Similar questions