# Homework Help: Newtons law Dynamics Question 4

1. May 27, 2013

### Nelonski

1. The problem statement, all variables and given/known data

Determine the acceleration of a 10kg block travelling up a ramp connected by a massless string and massless pulley to another block that is 5kg and in free-fall. if the coefficient of kinetic friction between the block and the ramp is 0.2

(The diagram shows one block that is 10kg travelling up a slope at 40 degrees, and a second block falling with a rope connected to it upwards)
2. Relevant equations
Newtons 2nd law

f_x = newtons law in x direction
f_y= newtons law in y direction
T= tension force
f_k = friction force
n= normal force
w_1 = weight of block 1 (ramp)
w_2= weight of block 2(ramp)
3. The attempt at a solution

FBD_1 : The FBD of the block going up the ramp has axis tilted so that the normal force is in the positive y direction, kinetic friction to the negative x direction, tension to the positive x direction, and weight which has two components to it pointing down.

FBD_2: The second block in free-fall has tension upwards since it is connected by a string and its weight downwards

Block on ramp: F_x= -f_k +T + w_1sin∅=ma (1)
F_y= n-w_1cos∅ (2)

Frictional force can be solved for by solving for normal force, then subbing into μkn.

The acceleration in block 1 and block two are the same with massless/frictionaless pulley and string approximation.

Block in free fall: F_y=T-w_2=m_2a (3)

.... The only unknown is the tension and the acceleration. So, I solved for the acceleration in
block 2 and subbed it into equation 1 to solve the tension. Then plugged the tension force into equation 1 again and solved for the acceleration..

Essentially, subbed equation 3 into equation 1, then plugged the tension value into eq 1 and solved for acceleration, which came out to be around 20 . seemed kind of high to me .

would those steps be considered right?

2. May 27, 2013

### Simon Bridge

... always include the units in any answer you produce: without units, it is impossible to tell if it is high or what?

You can check to see if it is high by figuring out the acceleration of the falling block if the block on the slope were not connected to it - then figuring if the effect of adding the slope block would make this acceleration faster or slower.

Your general approach is correct - if there is a problem, it will likely be in the details of the algebra or the arithmetic. eg - I think you may have the wrong direction for the x-component of weight in one of those equations.

3. May 27, 2013

### barryj

Weight = mass X g did you do this?

4. May 27, 2013

### barryj

Also, the tension will be the same for both FBDs. You can set the tension of one to be equal to the other and solv e for a. The answer might surprise you.

5. May 27, 2013

### Yuslek

OP here, yes that is what I did. The Tension in the 2nd FBD= Tension in first

and it was 21 m/s^2

To barryJ: Yes I used w=mg.

I don't see anything wrong with the Sol'n. I am quite confident in it, but if I did make some kind of mistake, then It would be appriciated if pointed out!

My final sol'n was to sub in : (T-w_2)/5=a into equation one, with a final equation of

(-f_k+wsin(theta)+10w_2)/5 = T

The tension came to be 153.47N.

Also the force of kinetic friction is actually 0.1 not 0.2 typo up there.

Last edited: May 27, 2013
6. May 28, 2013

### Simon Bridge

Welcome back Yoslek?
Did you reality check the acceleration as per post #2?

7. May 28, 2013

### Yuslek

Kind of, I dont really understand what he meant

Motion on a ramp without the block connected, then the block would be falling down at accelleration of gsin∅ right, which is 9.8sin40= 6.3m/s^2

So then the attatched the block should slow it down no? I dont really understand

Is the other block that is not on the ramp is just in freefall.. then it would just be accelleartion of 9.8m/s^2 right?.. And if a heavier block was attattched on a ramp then wouldnt it mean taht the block on the ramp would be falling down the ramp and the smaller block that is hanging be lifted upward?

8. May 28, 2013

### Simon Bridge

The block that is not on the ramp ... what would be it's acceleration if allowed to fall unattached?
Should it's acceleration when attached be bigger or smaller than that?
Is the acceleration you got bigger or smaller than that?
Therefore - does this support your intuition that the acceleration you got was too big?

9. May 28, 2013

### Yuslek

well if its not on the ramp and fell unattached then it would be free fall 9.8m/s^2 ... so when attached i think it should be going slower than that right? ... and might is much higher, so I guess the answer is wrong :S I dont see where it is wrong i checked so many times

I dont really know :/

I think I kind of see what I did wrong... since the block is sliding down then friction would be in the positive x direction, and tention would also be in the positive x direction because the block isnt moving up right?

So I have two vectors, friction and tension pointing positive x, with normal force up and weight force down with two components... so id have to redo all the math?

And the accelleration came out to be -3.73m/s^2... I think that makes a bit more sense.

Edit:: I recalculated it with these additions and tension in string became 30.237N.

Maybe that is right

Accelleration also came to be -3.73m/s^2

Newtons second law for the block on the ramp i made as F_x= T+f_k -w_1sin(theta)=ma

Because tension pulling the string in same direction as ramp, and friction is opposing movement, since the block is going down the ramp. are we just supposed to assume the heavier block pulls the entire system in its direction though?

Last edited: May 28, 2013
10. May 28, 2013

### Simon Bridge

Well done. Now you know it has to be wrong - so there must be a problem someplace.
You probably messed up the directions of the forces.

It is a good idea to check you have friction in the right direction - without friction, which way would the blocks slide? If the 10kg block is heavy enough to pull the 5kg one up, then you have friction in the wrong direction.

I'm going to define a shorthand to make it easier to write about things:

If M=10kg (is on the ramp) and m=5kg (the free fall one)
Then make sure that a positive movement of M makes a positive movement in m.

I think that you made "up" positive for m and up-the -slope positive for M as well - so a positive tension for M will be a negative tension for m. Did you do that?

It would have been better to choose down to be the +y direction for m.
That way the vectors will come out automatically.

Once that is done, pay careful attention to the directions of the forces in each diagram.

Watch the minus signs when you do the algebra.
Make sure you do all the algebra before substituting the values in.

The rest is arithmetic.

11. May 28, 2013

### Yuslek

I guess what I was basically saying in my calculation was that.. the tension in the string must be 153N inorder to pull the block up the ramp so friction is in the "negative" direction...

Yeah I think my recalculated one makes more sense..thanks