Question

A boy of mass $4kg$ is standing on a piece of wood with having a mass of $5kg$. The coefficient of friction between the wood and the floor is $0.5$. What is the maximum force that the boy can exert on the rope so that the piece of wood does not move from its place? (Round off to the Nearest Integer)

Answer 1

Step 1:Given Data

Mass of boy, $m_{Boy}=4kg$

Mass of wood, $m_{wood}=5kg$

Coefficient of static friction between boy and wood, $\mu =0.5$

Step 2: The formula used:

Newton's Second Law, $F_{net}=ma$ [where, $F_{net}$is the net force on the object, $a$ is the acceleration in SI units]

Step 3: Draw the free body diagram of the boy and the wooden box system, $\left(4+5\right)kg$

[Here, $T$ represents a tension in the rope, $f$ is the static frictional force, $N$ is the normal force on the body.]

Step 4: From the Free Body Diagram,

In the vertical direction, the net force $N+T=90$

$N=90-T$

In the horizontal direction, the net force,

$F_s=\mu N=T$

$\mu \left(90-T\right)=T$

$0.5(90-T)=T$

Substituting the values, the maximum force in the horizontal direction is found as

$90-T=2T$

$T=30N$

Hence, the maximum force in the horizontal direction is $30N$