Physics friction question (incline)
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Physics friction question (incline)

[From: ] [author: ] [Date: 12-02-14] [Hit: ]
5.explanation/equation appreciated.. ty :D~!-I will assume the rope pulls in a direction parallel to the ramp surface.F-net = 700 - 150(9.......
hi, im currently doing friction and need help...

A large crate (150kg) is being hauled up a 20m long 18° incline upon which µ = 0.17. The rope attached to the crate pulls it avec a force of 700N, and this is enough to give the crate a small acceleration up the incline.

Determine F-net on the crate and its acceleration (answer: 8.0N, 5.3 X 10^-2 m/s^2)

explanation/equation appreciated.. ty :D~!

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I will assume the rope pulls in a direction parallel to the ramp surface.
The friction force on the crate from the ramp is
Ff = μFn
Fn = mgcosθ
Ff = μmgcosθ

the portion of the weight tending to push the crate down slope is
F = mgsinθ

F-net = 700 - μmgcosθ - mgsinθ
F-net = 700 - mg(μcosθ + sinθ)
F-net = 700 - 150(9.81)(0.17cos18 + sin18)
►F-net = 7.37 N

F = ma
7.37 = 150a
►a = 4.9 X 10^-2 m/s^2

The difference in comparing answers to your givens is only in rounding errors in the values of sin18, cos18 and g. If the rope had been pulling at any angle other than parallel to the ramp, the given answers would have been smaller than they were, so that assumption is valid.

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One tip to further explain the work shown before me is change the coordinate system you use to define this system such that the x axis is parallel to the plane and the y is perpendicular just like this person did. This will help you get out of a lot of tedious sines and cosines since every force will be completely in the x or y axis with gravity being the exception. If you know about similar right triangles you can see that the component of gravity along the plane, or along the x-axis, is g*sin(18) why the perpendicular component, along the y, is g cos (18).
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