Force/Ramp problem

1. Nov 14, 2004

mrjeffy321

Here is the question:
I have attached the picture of the situation and below is how I tried to solve it.

Block A:
mass = 4 kg - given
the normal force = 34 N - normal force = m*g*cos(angle)
the force (parralell to the ramp surface) fulling it down
=19.62 N - force paralell = normal force * sin (angle)
no frictional force

Block B:
mass = 2 kg - given
normal force = 19.62 N - formal force = m*g
coefficent = kennetic friction = .5 - given
force of kinetic friction = 9.81 N - force of friction = normal force * coefficient of friction

so now there is a 19.62 N force pulling the blocks down, and a 9.81 N force trying the pull block B back, so as I see it, the tension in the cord is equal to 16.62-9.81 = 9.81 N. and then since there is a 9.81 N force acting on both blocks, the acceleration = net force / total mass = 1.64 m/s^2.
I got the acceleration correct, says the book, but not the tension force.

If I look in the backl of the book, it gives me the answer, which is not what I calculated.
A. Tension in the cord = 13 N
B. Acceleration = 1.6 m/s^2 ___I got this one right

I would really like to understand how to do this problem correctly.

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2. Nov 14, 2004

Gokul43201

Staff Emeritus
I can't see the attached picture yet (pending approval), but it looks like your approach is incorrect.

On block A (along direction of slope) : there's $Mgsin\theta$ acting downslope and T (tension) acting upslope. The resultant, $Mgsin\theta - T = Ma$

On block B : there's T pulling it towards the pulley, and F(fric) pulling in the opposite direction, so the net force here is $T - \mu mg = ma$

That's 2 equations in 2 unknowns.

3. Nov 14, 2004

mrjeffy321

I stil am not understanding.

woundnt the tension in the cord be equal to the diference in the forces acting on the objects (net force?)?

Can someone approve my picture that is attached.

4. Nov 14, 2004

Gokul43201

Staff Emeritus
No...that's not right.

You can not add or subtract forces acting on different objects. That's not how you find the net force. The net force is the (vector) sum of all the forces acting on a particular object.

Do you not follow the method described above ? It is merely using Newton's sencond Law on each of the 2 blocks..and it teaches you the correct way to calculate the net force.

I hope you are drawing free body diagrams for each block...else all this is in vain. You can't learn to do force problems without drawing free body diagrams.

5. Nov 14, 2004

mrjeffy321

I am drawing free body diagrams:

on block A, I have a force parallel to te ramp with a magnitude of 19.62 N, a normal force of 34 N, a force of weight of 39.24 N, and a force of tension that I know because I looked up the answer, but cant solve for of 13 N.

on block B, I have a wieght and normal force of 19.62 N, force of friction of 9.81 N, and a foce of tension that I know because I looked it up of 13 N.

6. Nov 14, 2004

Gokul43201

Staff Emeritus
Good !

Having resolved the weight (=39.2N) along the parallel (19.6N) and normal (43N) directions, you can now ignore the weight itself and consider only the components. So, along the parallel direction, you have 19.6 N forwards, tension (call it T) pulling backwards. The difference gives you the net force on this block in the parallel direction. This net force causes an acceleration (call it a) of this block along this direction. The force and acceleration are related by Newton's Second Law.

As of now, you don't know the tension yet (looking it up is not how you find it), but it's the same T as is on the other side of the rope. So, there's T pulling forward, and friction (=9.8N) pulling backwards. But you know that this block must accelerate forwards at the same rate (also=a) as the other block (since the rope is not elastic). So, the net force here is T - 9.8 which causes the acceleration = a. Write down this equation. (actually, I've already written it for you)

You now have 2 equations, and the unknowns are T and a. Solve them simultaneously to find T and a.

7. Nov 14, 2004

mrjeffy321

I dont know what you mean by,
"You now have 2 equations, and the unknowns are T and a. Solve them simultaneously to find T and a."
I just dont get it.

do you mean substitute one equalion into the other for m*a?

nothing I do will give me that magic number I am looking for, 13, which is a bad way to find it, but the only way I know right now.

8. Nov 15, 2004

Gokul43201

Staff Emeritus
I've actually written out the equations for you in post #2. If you substitute the values of M = 4kg and m = 2 kg, then the only unknowns in those equations are T and a.

To be able to solve any problem of this kind, you need to know how to solve 'simultaneous equations'. I suggest you learn that up.

In this case you can simply rearrange both equations so that there's only 'a' on the LHS and all the rest on the RHS.
$$a = blah1~~and~~a = blah2~~so,~blah1 = blah2$$
Solving this will give you T (I get T=13.07N). Now substitute this value of 'T' into any of the two equations to find the value of 'a'.

9. Nov 15, 2004

mrjeffy321

es, I understand now, it makes sense to me now, the force of tension is pulling it one way and frictiont the other way, which causes one acceleration, and for the other block, you have a parralell for one way and the tension the other, it dawned on me and is now very clear (at least in this example problem).
thanks