imported>Patrick at 16:52, 23 June 2011

2011-06-23T16:52:12Z

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Assume that the rope has a negligeable mass. The man (M) pulls on the rope (R) with a force <math>F_{MR} = 200</math> N towards the left while the rope pulls back on him with a force <math>F_{RM} = 200</math> N towards the right. Consequently, the tension in the rope is 200 N and the rope pulls on the box with a force <math>F_{RB} = 200</math> N towards left while the box pulls back on the rope with a force <math>F_{BR} = 200</math> N towards right. 
[[image:NIII_APP_EX_2_SOLNa.png|TOP]] 

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

The forces exerted on the box: <math>F_G</math> <math>_{EB}</math> (the gravitational force exerted on the box by the Earth), <math>F_N</math> <math>_{FB}</math> (the normal force exerted on the box by the floor), <math>F_{RB}</math> (the force exerted by the rope 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 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. 
[[image:NIII_APP_EX_2_SOLNb.png|TOP]] 

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_{RB}</math>
| -200 N
| 0
|}
 
Now we use Newton's Second Law: <math>\Sigma F_x = 0</math> (no acceleration)

<math>F_f</math> <math>_{FB} + (-200) = 0</math>

We find that the friction force has a magnitude of 200 N.
 
Then we use <math>\Sigma F_y = 0</math>

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

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: <math>F_G</math> <math>_{EM}</math> (the gravitational force exerted on the man by the Earth), <math>F_N</math> <math>_{FM}</math> (the normal force exerted on the man by the floor), <math>F_{RM}</math> (the force exerted on the man by the rope), <math>F_f</math> <math>_{FM}</math> (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. 
[[image:NIII_APP_EX_2_SOLNc.png|TOP]] 

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_{RM}</math>
| 200 N
| 0
|}
 
Now we use Newton's Second Law: <math>\Sigma F_x = 0</math> (no acceleration)

<math>-Ff</math> <math>_{FM} + 200 = 0</math>

We find that the friction force exerted on the man by the floor has a magnitude of 200 N.
 
Then we use <math>\Sigma F_y = 0</math>

<math>-700 + F_N</math> <math>_{FM} = 0</math>

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

Newton's 3rd Law EX 2 - Revision history

imported>Patrick at 16:52, 23 June 2011