Normal Force and Coefficient of Kinetic Friction

1. Sep 22, 2014

gcombina

A 250-N force is directed horizontally as shown to push a 29-kg box up an inclined plane at a constant speed. Determine the magnitude of the normal force, FN, and the coefficient of kinetic friction, µk.

So, I know I have to find the vertical and parallel forces
Vertical: Fn and 250 sin of angle (where is the angle symbol???)

so, if you look at the second picture, that is where I get stucked. Is "mg" also part of the parallel force?

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Last edited: Sep 22, 2014
2. Sep 22, 2014

dean barry

I think your normal force vector from the push force is upside down.
Split the weight of the block ( m * g ) into its parallel and normal vectors

3. Sep 23, 2014

gcombina

Thank you but I don't understand, what do you mean is upside down?

4. Sep 23, 2014

dean barry

Its easier if i sketch something up, ill get back to you.

5. Sep 23, 2014

nasu

The normal component of the pushing force is towards the plane. It's pushing the block against the incline, not trying to lift it from it.
Your arrow shows otherwise.

6. Sep 24, 2014

dean barry

heres that sketch, the upper shows the force resolution for the 250 N pushing force ( F1 )
F2 = F1 * cosine 27 °
F3 = F1 * sine 27 °
The lower shows the force resolution for the force of the block ( m * g ) F4
F5 = F4 * sine 27 °
F6 = F4 * cosine 27 °
F7 is the friction force and = ( F6 + F3 ) * µ
( µ = friction co-efficient )

Note : if the velocity is constant, forces up the slope = forces down the slope

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7. Sep 28, 2014

gcombina

thanks, let me study this. I just came back to the thread

8. Sep 28, 2014

gcombina

sorry but this confuses me more

9. Sep 28, 2014

CWatters

Which bit is confusing?

The 250N force acts towards the slope. The component that contributes to the normal force is F3
Gravity acts towards the slope. The component that contributes to the normal force is F6

The sum of (F3+F6) also act towards the slope.

You appear to be trying to make the trig show the sum pointing away from the slope? Perhaps you are confused because the normal force FN is usually shown as the equal and opposite reaction that ground makes on the block?

10. Sep 28, 2014

gcombina

11. Sep 28, 2014

haruspex

Because the first and third terms both act down the plane (positive x), while the second term acts up the plane.

12. Sep 28, 2014

CWatters

What Haruspex said.

With reference to Deans drawings and taking down the slope as positive..

translates as

F5 - F2 + F7 = 0

or you could instead take up the slope as positive, in which case that would be

-F5 + F3 - F7 = 0

13. Sep 29, 2014

dean barry

Why dont you work a fresh problem through first, (to get the feel) try this one :
A cannonball is launched at an angle of 40 ° above horizontal on level ground at a velocity of 500 m/s, calculate the maximum height reached, the time spent in the air and the horizontal range.
g = 9.81 (m/s)/s
Firstly split the launch velocity into its vertical and horizontal components.