Newtons Laws EX 2

First we focus on the mass $m_{2}=5$ kg. There are four forces acting on this particle: $F_{G}$, $F_{N}$, $F_{f}$, and $F_{T}$ pointing towards the string pulling the block. The particle moves down the incline and therefore the force of friction points up the incline. Similarly, we list forces on the mass $m_{1}=10$ kg. The force of tension pulls the block towards the left.

The blocks move at constant velocity and therefore their acceleration is zero. We select the x-axis parallel to the incline and parallel to the horizontal surface as shown in the diagram below.

We will deal with each particle separately. First we focus on the mass $m_{2}=2$ kg. The gravitational force points vertically down and therefore it makes a 30° angle with the negative y-axis; consequently, it makes an angle -120° with the positive x-axis.

The components of the forces exerted on $m_{2}$ are:

Forces	x-component	y-component
$F_{G}$	mg cos(-120°) = 50 cos(-120°) = -25 N	50 sin(-120°) = -43 N
$F_{N}$	0	$F_{N}$
$F_{f}$	$F_{f}$	0
$F_{T}$	$F_{T}$	0

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

First we have to deal with the y-component:

$-43+F_{N}=0$

This yields $F_{N}=43$ N

Now we can proceed to think about the x-component. Using that the acceleration is 0, we can write:

$F_{T}+F_{f}-25=0$

We cannot solve this equation immediately, and so, we focus on the mass $m_{1}=10$ kg.

The components of the forces exerted on $m_{1}$ are:

Forces	x-component	y-component
$F_{G}$	0	-100 N
$F_{N}$	0	$F_{N}$
$F_{f}$	15 N	0
$F_{T}$	$-F_{T}$	0

First we have to deal with the y-component:

$-100+F_{N}=0$

This yields $F_{N}=100$ N.

Now we can proceed to think about the x-component, and because acceleration is 0, we can write:

$-F_{T}+15=0$

There remains only to solve the following system of two equations:

$F_{T}+F_{f}-25=0$

$-F_{T}+15=0$

Solving for $F_{T}$, we find $F_{T}=15$ N and solving for $F_{f}$ we find $F_{f}=10$ N.