Newtons Laws EX 5

Firstly, we focus on the blue box of mass $m_1=18$ kg. The box is in contact with the surface of the incline and the surface of the box $m_2$. Therefore, there are two normal forces: $F_{N_1}$ which is exerted by the incline and $F_N$ which is exerted by the box $m_2$ on the box $m_1$. In addition, there is a gravitational force $F_G$ and the force $F$ that is exerted by Sally on $m_1$. Secondly, we list the forces exerted on the yellow box of mass $m_2=25$ kg. This box is in contact with the surface of the incline and the surface of the box $m_1$. Therefore, there is the normal forces $F_{N_2}$ exerted on this box by the incline and the normal force $F_N$ exerted by the box $m_1$. In addition, there is also the gravitational force $F_G$ exerted on it. It is important to note that the force $F$ is exerted by Sally only on the blue box $m_1$. Sally is not in contact with the yellow box $m_2$ and therefore she does not exert any force on it.

The blocks are at rest and therefore the sum of the forces on each box is zero. To draw a free body diagram for each of the boxes we select the x-axis parallel to the incline.

To determine the components of forces, we will begin with forces exerted on the blue box. The gravitational force makes and angle of -53° with the positive x-axis.

Forces	x-component	y-component
$F_G$	180 cos(-53°) = 108 N	180 sin(-53°) = -143 N
$F_{N_1}$	0	$F_{N_1}$
$F_N$	$F_N$	0
$F$	$-F$	0

Note that we used $g=10m/s^2$ in our calculations.

We deal first with the y-component:

$-143+F_{N_1}=0$

This yields $F_{N_1}=143$ N

Now we write the equation for the x-component:

$108+F_N-F=0$

We cannot solve this equation. First, we need to consider the yellow box.

The forces exerted on $m_2$ are $F_G$, the normal force exerted by the blue box $F_N$ and the normal force exerted by the incline $F_{N_2}$. The gravitational force makes the angle of -53° with the positive x-axis.

The components of the forces exerted on $m_2$ are:

Forces	x-component	y-component
$F_G$	250 cos(-53°) = 150 N	250 sin(-53°) = -200 N
$F_{N_2}$	0	$F_{N_2}$
$F_N$	$-F_N$	0

We write the equation for the y-component:

$-200+F_{N_2}=0$

This gives us $F_{N_2}=200$ N.

The equation for the x-component yields:

$150-F_N=0$

We have two equations with two unknowns:

$108+F_N-F=0$

$150-F_N=0$

By adding the two equations we solve $F=258$ N.

We solve the second equation for $F_N=150$ N.