Newtons Laws EX 4

First, we list the forces acting on $m_{1}=4$ kg. There are four forces acting on this particle: $F_{G}$, $F_{N}$, $F_{f}$ (pointing down the incline because the block moves up the incline), $F_{T}$ (pointing towards the string pulling the block).

We also list forces acting $m_{2}=8$ kg. There are five forces acting on this particle: $F$, $F_{G}$, $F_{N}$, $F_{f}$ (pointing to the right because the particle moves to the left), $F_{T}$ (pointing towards the string pulling the block).

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 focus on the mass $m_{1}=4$ kg. The gravitational force makes an angle of -53° with the positive x-axis. The components of the forces exerted on $m_{1}$ are:

Forces	x-component	y-component
$F_{G}$	40 cos(-53°) = 24 N	49 sin(-53°) = -32 N
$F_{N}$	0	$F_{N}$
$F_{f}$	20 N	0
$F_{T}$	$-F_{T}$	0

First we have to deal with the y-component:

$-32+F_{N}=0$

This yields $F_{N}=32$ N.

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

$-F_{T}+20+24=0$

We solve $F_{T}=44$ N

We then focus on $m_{2}=8$ kg. The force F makes an angle of 150° with the positive x-axis. The components of the forces exerted on $m_{2}$ are:

Forces	x-component	y-component
$F_{G}$	0	-80 N
$F_{N}$	0	$F_{N}$
$F_{f}$	20 N	0
$F_{T}$	44 N	0
$F$	F cos(150°) = -0.87 F	F sin(150°) = 0.5 F

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

First we may write the equation for the y-component:

$-80+F_{N}+0.5F=0$

We cannot solve this equation yet.

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

$44+20-0.87F=0$

Then solving for F, we find $F=74$ N

Now we solve for $F_{N}$:

$F_{N}=43$ N