Newton's 3rd Law EX 2

Assume that the rope has a negligeable mass. The man (M) pulls on the rope (R) with a force $F_{MR}=200$ N towards the left while the rope pulls back on him with a force $F_{RM}=200$ N towards the right. Consequently, the tension in the rope is 200 N and the rope pulls on the box with a force $F_{RB}=200$ N towards left while the box pulls back on the rope with a force $F_{BR}=200$ N towards right.

(a) First we will consider the forces exerted on the box.

The forces exerted on the box: $F_{G}$ $_{EB}$ (the gravitational force exerted on the box by the Earth), $F_{N}$ $_{FB}$ (the normal force exerted on the box by the floor), $F_{RB}$ (the force exerted by the rope on the box), $F_{f}$ $_{FB}$ (the frictional force exerted by the floor on the box).

The box does not move and therefore its acceleration is zero. We will choose the coordinate system with x-axis parallel to the floor. The diagram below shows all the forces acting on the box and the appropriate free body diagram.

The components of these forces are shown in the table below:

Forces	x-component	y-component
$F_{G}$ $_{EB}$	0	$-mg=-(100\times 10)=-1000$ N
$F_{N}$ $_{FB}$	0	$F_{N}$ $_{FB}$
$F_{f}$ $_{FB}$	$F_{f}$ $_{FB}$	0
$F_{RB}$	-200 N	0

Now we use Newton's Second Law: $\Sigma F_{x}=0$ (no acceleration)

$F_{f}$ $_{FB}+(-200)=0$

We find that the friction force has a magnitude of 200 N.

Then we use $\Sigma F_{y}=0$

$-1000+F_{N}$ $_{FB}=0$

We find that the normal force has a magnitude of 1000 N.

(b) Consider the forces exerted on the man.

The forces exerted on the man: $F_{G}$ $_{EM}$ (the gravitational force exerted on the man by the Earth), $F_{N}$ $_{FM}$ (the normal force exerted on the man by the floor), $F_{RM}$ (the force exerted on the man by the rope), $F_{f}$ $_{FM}$ (the frictional force exerted on the man by the floor).

The man does not move and therefore his acceleration is zero. We will choose the coordinate system with x-axis parallel to the floor. The diagram below shows all the forces acting on the man and the appropriate free body diagram.

The components of these forces are shown in the table below:

Forces	x-component	y-component
$F_{G}$ $_{EM}$	0	$-mg=-(70\times 10)=-700$ N
$F_{N}$ $_{FM}$	0	$F_{N}$ $_{FM}$
$F_{f}$ $_{FM}$	$-F_{f}$ $_{FM}$	0
$F_{RM}$	200 N	0

Now we use Newton's Second Law: $\Sigma F_{x}=0$ (no acceleration)

$-Ff$ $_{FM}+200=0$

We find that the friction force exerted on the man by the floor has a magnitude of 200 N.

Then we use $\Sigma F_{y}=0$

$-700+F_{N}$ $_{FM}=0$

We find that the normal force exerted on the man by the floor has a magnitude of 700 N.