imported>Patrick at 18:17, 30 August 2011

2011-08-30T18:17:34Z

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First we focus on the mass <math>m_2 = 5</math> kg. There are four forces acting on this particle: <math>F_G</math>, <math>F_N</math>, <math>F_f</math>, and <math>F_T</math> 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 <math>m_1 = 10</math> kg. The force of tension pulls the block towards the left. 
[[image:N_APP_EX_2_SOLNa.png|TOP]] 

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. 
[[image:N_APP_EX_2_SOLNb.png|TOP]] 

We will deal with each particle separately. First we focus on the mass <math>m_2 = 2</math> 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 <math>m_2</math> are:

{| border="1" cellpadding="2" style="text-align:center"
|-
! Forces
! x-component
! y-component
|-
! <math>F_G</math>
| mg cos(-120°) = 50 cos(-120°) = -25 N
| 50 sin(-120°) = -43 N
|-
! <math>F_N</math>
| 0
| <math>F_N</math>
|-
! <math>F_f</math>
| <math>F_f</math>
| 0
|-
! <math>F_T</math>
| <math>F_T</math>
| 0
|}

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

First we have to deal with the y-component:

<math>-43 + F_N = 0</math>

This yields <math>F_N = 43</math> N
 
Now we can proceed to think about the x-component. Using that the acceleration is 0, we can write:

<math>F_T + F_f - 25 = 0</math>

We cannot solve this equation immediately, and so, we focus on the mass <math>m_1 = 10</math> kg.
 
The components of the forces exerted on <math>m_1</math> are:

{| border="1" cellpadding="2" style="text-align:center"
|-
! Forces
! x-component
! y-component
|-
! <math>F_G</math>
| 0
| -100 N
|-
! <math>F_N</math>
| 0
| <math>F_N</math>
|-
! <math>F_f</math>
| 15 N
| 0
|-
! <math>F_T</math>
| <math>-F_T</math>
| 0
|}

First we have to deal with the y-component:

<math>-100 + F_N = 0</math>

This yields <math>F_N = 100</math> N.
 
Now we can proceed to think about the x-component, and because acceleration is 0, we can write:

<math>-F_T + 15 = 0</math>
 
There remains only to solve the following system of two equations:

<math>F_T + F_f - 25 = 0</math>

<math>-F_T + 15 = 0</math>

Solving for <math>F_T</math>, we find <math>F_T = 15</math> N and solving for <math>F_f</math> we find <math>F_f = 10</math> N.

Newtons Laws EX 2 - Revision history

imported>Patrick at 18:17, 30 August 2011