How Do You Calculate Friction Force on an Inclined Ramp?

AI Thread Summary
To calculate the friction force on an inclined ramp, the angle of incline is crucial, which can be determined using the sine function (sin theta = height/ramp length). The work done by friction is given as -0.06, and the relationship between work, friction force, and displacement involves the cosine of the angle between the displacement and friction force vectors. Since friction opposes motion, the angle should be considered as 180 degrees when calculating work done by friction. Clarification on the angle of incline is essential for accurate calculations, and using both sine and cosine relationships can help determine the coefficient of friction.
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Homework Statement



A cart slides down an inclined ramp at is 1.44m long. The height of the ramp is .10m. The angle of incline of the ramp is not given. The work done by the friction is -.06. I need to find the friction force.

Homework Equations



Work done by the friction = Friction force * d cos theta.

The Attempt at a Solution



When looking through my books, I found what appears to be two contradictory ways of determining theta - 1) by using sin (sin theta = .10/1.44) so the angle of incline is 4 or 2) by using 180 for theta because the friction force is acting in the opposite direction. Can anyone point me in the right direction? Thanks for your help.
 
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W_{Friction}=Fd\cos \theta

Here, theta is the angle between the displacement vector, d and the friction force vector.

What direction does d point in?

What direction does the friction vector point in?If you can answer the above equations, you can answer the following:
What is the angle between them?

Hopefully, this helps lead you on the right path. See if you can solve this now.
 
So, as the friction vector is traveling in the opposite direction of the displacement vector, then I should go with cos 180.

Earlier, I used the angle of incline (sin theta = .10/1.44) to determine the coefficient of friction of a block on the ramp (e.g. in mg sin theta & mg cos theta) Was this correct?
 
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