imported>Patrick at 02:44, 15 April 2011

2011-04-15T02:44:25Z

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

[[image:N_APP_EX_5_SOLNa.png|TOP]] 

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.

[[image:N_APP_EX_5_SOLNb.png|TOP]] 

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.

{| border="1" cellpadding="2" style="text-align:center"
|-
! Forces
! x-component
! y-component
|-
! <math>F_G</math>
| 180 cos(-53°) = 108 N
| 180 sin(-53°) = -143 N
|-
! <math>F_{N_1}</math>
| 0
| <math>F_{N_1}</math>
|-
! <math>F_N</math>
| <math>F_N</math>
| 0
|-
! <math>F</math>
| <math>-F</math>
| 0
|}

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

We deal first with the y-component:

<math>-143 + F_{N_1} = 0</math>

This yields <math>F_{N_1} = 143</math> N
 
Now we write the equation for the x-component:

<math>108 + F_N - F = 0</math>

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

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

The components of the forces exerted on <math>m_2</math> are:

{| border="1" cellpadding="2" style="text-align:center"
|-
! Forces
! x-component
! y-component
|-
! <math>F_G</math>
| 250 cos(-53°) = 150 N
| 250 sin(-53°) = -200 N
|-
! <math>F_{N_2}</math>
| 0
| <math>F_{N_2}</math>
|-
! <math>F_N</math>
| <math>-F_N</math>
| 0
|}

We write the equation for the y-component:

<math>-200 + F_{N_2} = 0</math>

This gives us <math>F_{N_2} = 200</math> N.
 
The equation for the x-component yields:

<math>150 - F_N = 0</math>
 

We have two equations with two unknowns:

<math>108 + F_N - F = 0</math>

<math>150 - F_N = 0</math>

By adding the two equations we solve <math>F = 258</math> N.

We solve the second equation for <math>F_N = 150</math> N.

Newtons Laws EX 5 - Revision history

imported>Patrick at 02:44, 15 April 2011