# Inclined Plane block of mass

1. Nov 22, 2005

A block with mass 10.0 kg is placed on an inclined plane with slope angle 30 degrees and is connected to a second hanging block that has mass m by a cord passing over a small, frictionless pulley. The coefficient of static friction is 0.45, and the coefficient of kinetic friction is 0.35.
(a) Find the mass m for which the 10.0 kg block moves up the plane at a constant speed once it has been set in motion.
(b) Find the mass m for which it moves down the plane at constant speed once it has been set in motion.
(c) For what range of values of m will the block remain at rest if it is released from rest? (Use g = 10 m/s^2)

So the first thing I note is that $$\mu_{s} = 0.45$$ and $$\mu_{k} = 0.35$$. I also know that I am dealing with a 30-60-90 right triangle, which will make the problem easier. Now I draw the free body diagram. The force of gravity $$w = mg$$ is acting on the both blocks. The tension of the rope is acting on both blocks(same for each block). Static friction is acting on the 10.0 kg block, directed in the $$- x$$ direction.

So for part (a) the key phrase is once it has been set in motion . That means kinetic friction is important in this case. So I calculate the kinetic friction of the block to be: $$f_{k} = (0.35)(100 N) = 35 N$$. What would I do from here?

For part (b), the mass has to be less than the mass of the block from the previous question, because it is moving down the ramp.

For (c) the masses have to be such that they do not break overcome the static friction $$f_{s}$$.

Any help is appreciated

Thanks

2. Nov 22, 2005

### mezarashi

For (a), now that you have the force for friction, what other forces are acting on the block? And in which direction? If velocity is constant, so there is no acceleration, thus the net force must be zero.

For (b), intuition may be failing you here. If it is moving down the ramp at constant velocity, what is its acceleration? Consequently what must its net force be?

3. Nov 22, 2005

### marlon

Let us first (as we should always do) write down the equations of motion (Newton's Second Law).

1) pick x axis (along the incline) and an y axis (perpendicular to the incline)
2) as a convention, we agree that forces direction upward the incline have a positive sign and the forces with opposite direction have - sign.
3) if you look at block 1 on the incline, which forces are acting on it. There are 4 of them. For each force, calculate both x and y components
4) for block 2, again which forces are acting on it ? In this case it is just gravity...

Once you have solved these 4 question, as you should LAWAYS do with these problems, you can actually start solving your questions

marlon

EDIT/ Besides, the friction force you calculated is wrong. Do the first 4 steps as i outlines and you will se why. It has something to do with the fact that in the y direction (along which the normal force is defined), gravity is NOT g !!!

Last edited: Nov 22, 2005
4. Nov 22, 2005

1. Picked x-axis along incline and y-axis perpendicular to incline
2. Agreed
3. The forces acting on block 1 are its weight ( 100 N ), the tension of the rope T , the normal force n , and the static friction force $$f_{s}$$. The x and y components of the weight are, $$F_{gx} = mg\sin 30$$, $$F_{gy} = mg\cos 30$$. $$f_{sx} = -100\cos 30(0.45)$$ and there is no y-component. The normal force has no x-component. The weight of the second block has $$mg$$ as its y component.

Last edited: Nov 22, 2005
5. Nov 22, 2005

### marlon

By 0 you mean 30, right ? I am sure that is just a typo.

Along x : $$m_1a_{1x} = T - \mu_kN -mgsin(30°)$$
Along y : $$m_1a_{1y} = 0 = -mgcos(30°) + N$$ Why can i write 0 here ?

For mass two there is only an y component:
$$m_2a_{2y} = T-m_2g$$,

These are the equations of motion in their most general form.
Now, look at what has been asked and see what you can do with the accelerations. I mean, can they get 0 ? is $$a_1 (= \sqrt{a_{1x}^2 + a_{1y}^2})=a_2$$ ? Be sure to examine this for each question...

You see, this is how you should always proceed

marlon

Last edited: Nov 22, 2005
6. Nov 22, 2005

So $$m_1a_{1x} = T - \mu_kN -mgsin(30°)$$ you set $$a_{1x} = 0$$ and solve for m ? And the tension would just be equal to the weight of the second block. You use $$f_{k}$$.

Same thing for part B except you reverse the signs?

And For part (x) you just solve for m again?

7. Nov 23, 2005

Correct

Correct

marlon