# Friction problems

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rock.freak667
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## The Attempt at a Solution

i've drawn a free body diagram and got my equations to be
x=-490sin30 + f=50a
y=-490cos30 + n = 50a
and now im stuck

the 'x' one is correct.

For the 'y' one, the resultant force perpendicular to the plane is 0 so -490cos30+N = ? (so what is N?)

n is

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rl.bhat
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Hi zhangsterii, welcome to PF.
Since crate is moving down a should be...... ?
In the y direction there is no motion. So in that direction acceleration is zero.
Now what is the frictional force f?

[quote=

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rl.bhat
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For both x and y you have written 50a. How is it that?

JONNYWOorking on part 2

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how should the x= and y= equations be set up??

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rl.bhat
Homework Helper
When the particle will have uniform velocity in the upward direction?

When the particle will have uniform velocity in the upward direction?
umm wat are u talking about?

weeee

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rl.bhat
Homework Helper
I am asking "when the particle moves with uniform velocity?" Refer Newtons first law.

when there is no accerlation

rl.bhat
Homework Helper
Correct.
When the acceleration is zero on an inclined plane with friction?

Correct.
When the acceleration is zero on an inclined plane with friction?

when velocity is constant

fagito

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rl.bhat
Homework Helper
Now If you pull the crate in the upward direction, what is the direction of the frictional force?
What is the net down ward force?
To move the crate with uniform velocity, how much force you have to apply?

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hi joonwoo

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rl.bhat
Homework Helper
the direction of friction would be downward and to move the crate, you would need a force equal to the force of friction?
Along with the frictional force you have to overcome the component of weight in that direction to pull the crate up.