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Inclined Plane with block of mass

  • Thread starter danago
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  • #1
danago
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[SOLVED] Inclined Plane

A block of mass m sits on an inclined plane. The coefficient of kinetic friction is known. If the tension in the rope is T, and at the instant shown the block is moving down the slope, what is the blocks acceleration up the slope?

http://img374.imageshack.us/img374/2218/36905063xh5.gif [Broken]

I started by drawing a free body diagram of the block.

http://img111.imageshack.us/img111/8047/96925869yv4.jpg [Broken]

Where W is the force due to the weight of the block, P is the force from the pulley, F is the friction force and N is the normal force.

I can easily calculate the weight force and resolve it into components down the slope and perpendicular to the slope to find the normal force and the force down the slope. I can then use the normal force to find the frictional force.

What i am having trouble with is finding the force due to the pulley. I tried drawing a FBD of it, but im not really sure how the tension in the rope applies a force to the pulley. Do i assume the force to be applied to the point where the rope first contacts each side of the pulley?
 
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Answers and Replies

  • #2
danago
Gold Member
1,122
4
I came up with the following forces acting on the block:

[tex]
\begin{array}{l}
\overrightarrow W = \left( {\begin{array}{*{20}c}
{mg\sin \theta } \\
{ - mg\cos \theta } \\
\end{array}} \right) \\
\overrightarrow N = - \overrightarrow {W_y } = \left( {\begin{array}{*{20}c}
0 \\
{mg\cos \theta } \\
\end{array}} \right) \\
\overrightarrow F = - \mu _k \left| {\overrightarrow N } \right|\widehat{\underline i } = \left( {\begin{array}{*{20}c}
{ - \mu _k mg\cos \theta } \\
0 \\
\end{array}} \right) \\
\end{array}
[/tex]

The notation i am using for my vectors is as follows:

[tex]
\left( {\begin{array}{*{20}c}
a \\
b \\
\end{array}} \right) = a\widehat{\underline i } + b\widehat{\underline j }
[/tex]

Where i is a unit vector down the slope and j is a unit vector normal to the surface of the slope.
 
  • #3
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assume the angle between ropes constant and the tension on rope is T.

P = T + T Cos (alpha)
(P is the force you show above)

N = mg Cos (alpha) - T Sin (alpha)
 
  • #4
danago
Gold Member
1,122
4
Oh yep i see what i did wrong now. I forgot to consider the effect of the pulley on the normal force, which affected the friction force i calculated.

Thanks for the help :smile:
 

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