Question

If x is the coefficient of friction between the board and floor as shown in figure. Find the maximum force that man can exert on the rope, so that the board does not slip on the floor

Answer 1

Free Body Diagram of the board is shown below.

$\setlength{\unitlength}{1mm}\begin{picture}(5,5)\put(0,0){\circle*{2}}\put(0,0){\vector(1,0){20}}\put(0,0){\vector(-1,0){10}}\put(0,0){\vector(0,-1){20}}\put(0,0){\vector(0,1){20}}\put(-8,-24){ \sf{mg+R_m} }\put(-2,22){ \sf{R_f} }\put(21,-1){ \sf{T} }\put(-21,-1){ \sf{f_f+f_m} }\put(1.5,-3.5){\sf{m}}\end{picture}$

• The board experiences weight $\sf{mg}$ downward.

• Reaction force due to floor $\sf{R_f}$ acts on the board upward.

• The man is standing over the board so the reaction force due to him $\sf{R_m}$ acts on the board downward.

• The tension in the string $\sf{T}$ is acted upon the board rightward.

• Due to the tension in the string, the board tends to move rightward, and so the board experiences frictional force due to the floor $\sf{f_f}$ and that due to the man $\sf{f_m}$ leftward.

Since net vertical force is zero,

$\sf{\longrightarrow R_f=R_m+mg}$

As the board does not slip over the floor, net horizontal force should be zero. Then,

$\sf{\longrightarrow T=f_f+f_m}$

Since coefficient of friction between board and floor is $\sf{x,}$

$\sf{\longrightarrow T=x\cdot R_f+f_m}$

Putting value of $\sf{R_f,}$

$\sf{\longrightarrow T=x(R_m+mg)+f_m\quad\quad\dots(1)}$

Free Body Diagram of the man is shown below.

$\setlength{\unitlength}{1mm}\begin{picture}(5,5)\put(0,0){\circle*{2}}\put(0,0){\vector(0,-1){20}}\put(0,0){\vector(0,1){20}}\put(0,0){\vector(1,0){10}}\put(-3,-25){ \sf{Mg} }\put(-6,23){ \sf{R_m+T} }\put(11.5,-1){ \sf{f_m} }\end{picture}$

• The man experiences weight $\sf{Mg}$ downward.

• The reaction due to the board is the same as $\sf{R_m}$ and it acts on the man upward.

• The man pulls the rope with a maximum force $\sf{T}$ (to be found) downward, so the man experiences an equal reaction $\sf{T}$ according to Newtons' Third Law in upward direction.

• As the board tends to move rightward, the man tends to move leftward due to inertia, and so he experiences frictional force due to the board, equal to $\sf{f_m,}$ in rightward direction.

Since net vertical force is zero,

$\sf{\longrightarrow Mg=R_m+T}$

$\sf{\longrightarrow R_m=Mg-T}$

Since the board does not slip over the floor, the man won't move over the board. Hence,

$\sf{\longrightarrow f_m=0}$

Then (1) becomes,

$\sf{\longrightarrow T=x(Mg-T+mg)}$

$\sf{\longrightarrow T=Mgx-Tx+mgx}$

$\sf{\longrightarrow T+Tx=Mgx+mgx}$

$\sf{\longrightarrow T(1+x)=(M+m)gx}$

$\sf{\longrightarrow\underline{\underline{T=\dfrac{(M+m)gx}{1+x}}}}$

This is the maximum force the man can exert on the rope.

$\sf{\longrightarrow\underline{\underline{F_{max}=\dfrac{(M+m)gx}{1+x}}}}$