Question

The friction coefficient between the board and the floor shown in figure (6−E7) is μ. Find the maximum force that the man can exert on the rope so that the board does not slip on the floor.
Figure

Answer 1

Answer:

Solution :

Let the maximum force exerted by the man is T.

From the free body diagram 

R+T=Mg

R=Mg-T-------------(1)

Again R1-R-mg=0

R1=R+mg----------------(2)

and T-μR1=0

From equation (ii), T-μ(R+mg)=0

T-μ(Mg +T) -μmg=0

T(1+μ)=μMg+μmg=0

T(1+μ)=μ(M+m)g

T=μ(M+m)g/1+μ

Maximum Fore exerted by man is μ(M+m)g/1+μ

Answer 2

The maximum force exercised by the man is $\mu M+\mu+m g 1$

Explanation:

Let the maximum force applied on the rope by the man be denoted as T.

From the diagram,

$T+R=M g$

$\Rightarrow R=M g-T$                    (1)

Again,

$\Rightarrow R 1=R+m g$                (2)

and $-\mu R 1+T=0$

From Equation (2),

$T-\mu(R+m g)=0$

$T+\mu t-\mu M g-\mu m g=0$

$\Rightarrow \mu M g+\mu m g=T(1+\mu)$

$\Rightarrow T=\mu M+\mu+m g 1$

So, the maximum force exercised by the man is $\mu M+\mu+m g 1$.