View source for Newton's 3rd Law EX 3

The man pushes on the box with force of <math>F_{MB} = 200</math> N at 30° below the horizontal while the box pushes back on him with an equal and opposite force <math>F_{BM} = 200</math> N at 30° above the horizontal.<br><br>
[[image:NIII_APP_EX_3_SOLNa.png|TOP]]<br><br>

'''(a)''' The system we will consider now is the box.

The forces exerted on the box: <math>F_G</math> <math>_{EB}</math> (the gravitational force exerted by the Earth on the box), <math>F_N</math> <math>_{FB}</math> (the normal force exerted by the floor on the box), <math>F_{MB}</math> (the force exerted by the man on the box), <math>F_f</math> <math>_{FB}</math> (the frictional force exerted by the floor on the box).

The box does not move and therefore the 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.<br><br>
[[image:NIII_APP_EX_3_SOLNb.png|TOP]]<br><br>

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

{| border="1" cellpadding="2" style="text-align:center"
|-
! Forces
! x-component
! y-component
|-
! <math>F_G</math> <math>_{EB}</math>
| 0
| <math>-mg = -(100 \times 10) = -1000</math> N
|-
! <math>F_N</math> <math>_{FB}</math>
| 0
| <math>F_N</math> <math>_{FB}</math>
|-
! <math>F_f</math> <math>_{FB}</math>
| <math>-F_f</math> <math>_{FB}</math>
| 0
|-
! <math>F_{MB}</math>
| <math>F_{MB} \cos(-30^{\circ}) = (200)(0.867) = 173</math> N
| <math>F_{MB} \sin(-30^{\circ}) = (200)(-0.5) = -100</math> N
|}
<br>
Now we use Newton's Second Law: <math>\Sigma F_x = 0</math> (no acceleration)

<math>-F_f</math> <math>_{FB} + 173 = 0</math>

We find that the friction force has a magnitude of 173 N.
<br><br>
Then we use <math>\Sigma F_y = 0</math>

<math>-1000 + F_N</math> <math>_{FB} + (-100) = 0</math>

We find that the normal force has a magnitude of 1100 N.<br><br>

'''(b)''' The system we will consider now is the man.

The forces exerted on the man: <math>F_G</math> <math>_{EM}</math> (the gravitational force exerted by the Earth on the man), <math>F_N</math> <math>_{FM}</math> (the normal force exerted by the floor on the man), <math>F_{BM}</math> (the force exerted by the box on the man), <math>F_f</math> <math>_{FM}</math> (the frictional force exerted by the floor on the man).

The man does not move and therefore the 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.<br><br>
[[image:NIII_APP_EX_3_SOLNc.png|TOP]]<br><br>

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

{| border="1" cellpadding="2" style="text-align:center"
|-
! Forces
! x-component
! y-component
|-
! <math>F_G</math> <math>_{EM}</math>
| 0
| <math>-mg = -(70 \times 10) = -700</math> N
|-
! <math>F_N</math> <math>_{FM}</math>
| 0
| <math>F_N</math> <math>_{FM}</math>
|-
! <math>F_f</math> <math>_{FM}</math>
| <math>-F_f</math> <math>_{FM}</math>
| 0
|-
! <math>F_{BM}</math>
| <math>F_{BM} \cos(150^{\circ}) = (200)(- 0.867) = -173</math> N
| <math>F_{BM} \sin(150^{\circ}) = (200)(0.5) = 100</math> N
|}
<br>
Now we use Newton's Second Law: <math>\Sigma F_x = 0</math> (no acceleration)

<math>F_f</math> <math>_{FM} + (-173) = 0</math>

We find that the friction force has a magnitude of 173 N.
<br><br>
Then we use <math>\Sigma F_y = 0</math>

<math>-700 + F_N</math> <math>_{FM} + 100 = 600</math>

We find that the normal force has a magnitude of 600 N.