imported>Patrick at 00:36, 18 June 2011

2011-06-18T00:36:39Z

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There are four forces acting on this block: <math>F_G</math>, <math>F_N</math>, <math>F_f</math>, <math>F</math>. The block slides up the incline and therefore the force of friction points down the incline (because friction opposes the slide between the two surfaces). 
[[image: Helena Stat Dyn Sol 2.png|TOP]] 

We select the x-axis parallel to the incline and then draw the free body diagram. 
[[image: Helena Stat Dyn Sol 2b.png|TOP]] 

The components of the forces exerted on the mass <math>m = 5</math> kg are:

{| border="1" cellpadding="2" style="text-align:center"
|-
! Forces
! x-component
! y-component
|-
! <math>F_G</math>
| 50 cos(-127°) = -30 N
| 50 sin(-127°) = -40 N
|-
! <math>F_N</math>
| 0
| <math>F_N</math>
|-
! <math>F_f</math>
| <math>-F_f</math>
| 0
|-
! <math>F</math>
| 40 cos(-37°) = 32 N
| 40 sin(-37°) = -24 N
|}
 
Note that we used <math>g = 10 m/s^2</math> in our computations.
 
First we have to deal with the y-component.

Writing

<math>-40 - 24 + F_N = 0</math>

we see that <math>F_N = 64</math> N.
 
Now, looking at the x-component, we write

<math>-30 + 32 - F_f = 0</math>

solving for <math>F_f</math> we find <math>F_f = 2</math> N.

Static and Dynamic Equilibrium EX 2 - Revision history

imported>Patrick at 00:36, 18 June 2011