# Homework Help: Find the speed of the brick.

1. Feb 24, 2005

### sony

"A brick (M) is connected to another brick on a 30 degree slope. Don't mind the weight of the rope connecting these, and the friction.

The mass of the right brick (M) is 3,5 kg

What is the speed of the brick (M) when it have falled 0,5m?"

I've drawed forces on the illustration.

Here is a picture of what I've done (I wrote it in mathtype, I don't know how to paste it here)
http://home.no.net/erfr1/images/1.gif

I just don't understand how to find the acceleration without knowing both masses.

Thank you for your help!

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2. Feb 24, 2005

### NateTG

Are both bricks on the slope?

3. Feb 24, 2005

### sony

The right brick is hanging in the air, connected to the left brick by thin rope.

4. Feb 24, 2005

### sony

Maybe slope wasn't the right word to use, I should have checked the dictionary. Isn't the illustration clear?

5. Feb 24, 2005

### NateTG

I can't see the picture - it's waiting for approval.

It really seems like youre missing something then - perhaps the bricks should have the same mass?

6. Feb 24, 2005

### sony

If I use the same mass I don't get the correct answer (which is 1,9 m/s). The answer doensn't say anything about assumptions (friction, mass), so I guess you can rule the first mass out some way. I have no idea how though.

7. Feb 24, 2005

### NateTG

I'm going to have to wait for the picture before I try anything more. Sorry that I couldn't help.

8. Feb 24, 2005

### HallsofIvy

The rope from the brick on the slope (yes, that's the right word) goes over a pulley and then is attached to a brick hanging vertically.

There are, then, two forces acting on the bricks. There is the force of gravity, mg= (3,5 kg)(9,8 m/s) acting on the right brick. There is also the component of gravity down the slope, acting on the left brick. You don't say what the mass of the left brick is but assuming it is also 3,5 kg, That is (3,5)(9,8)cos(30). The pulley causes those two forces to be opposing: the net force is (3,5)(9,8)(1- cos(30)) up the slope. What acceleration will that give the total 7,0 kg mass?

9. Feb 25, 2005

### sony

Thank you both for answers.

I'm afraid I dont fully understand. Isn't it the x-component (horizontal comp.) of gravity that should be used? I get that Gx on the left brick is mg*sin(30) down the slope. And were does (1-cos(30)) come from? That is the pull upward right? "S" on the drawing? I thought the that S was the weight of the right brick (mg)?

What does it mean that "The pulley causes those two forces to be opposing"?

Thanks

10. Feb 25, 2005

### HallsofIvy

Gravity has NO horizontal component!

Assuming that the 30 degrees is "angle of elevation", measured with the horizontal, then the force of gravity makes a 60 degree angle with the slope and the force down the slope is mg sin(60)= mg cos(30), not mg sin(30).

The S on the drawing appears to me to be the tension in the string- the force pulling up on the left brick- and is the same as S1, the tension in the vertical portion. It certainly is not the weight of the brick. The right brick is being pulled downward by the force of gravity, mg, and upward by the tension in the string, S1= S. The left brick is being pulled down the slope by the gravitational force down the slope, mg cos(30), and up the slope by the tension in the string, S1= S. The pulley changes the direction of the forces- we can think of this as one long line with those four forces acting along that line. Since the two "tensions" cancel, we can treat this as a single object acted on by force mg- mg cos(30)= mg(1- cos(30) (assuming the two bricks have the same mass). Since the mass of the whole system is 2m, we have 2m(acc)= mg(1- cos(30)) or acceleration= (g/2)(1- cos(30)).

11. Feb 25, 2005

### sony

Ok, I feel pretty stupid here. I still don't fully get it.

"Assuming that the 30 degrees is "angle of elevation", measured with the horizontal, then the force of gravity makes a 60 degree angle with the slope and the force down the slope is mg sin(60)= mg cos(30), not mg sin(30)."
-> Is this Gx drawn on the left brick?

"mg(1- cos(30)" Is this "S" on the drawing?

I hate to ask, but could you maybe upload an illustration of what you have done?

If not I understand perfectly. I can wait till we have class this monday, so you don't have to waste more time on me :). Maybe we have another curriculum, so my physics teacher could explain.

Thanks anyway.

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