# One block pushing into another

1. Oct 8, 2008

### zoner7

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

Two cubic blocks are in contact, resting on a frictionless horizontal surface. The block on the left has a mass of mL = 6.70 kg, and the block on the right has a mass of mR = 18.4 kg. A force of magnitude 112 N is applied to the left face of the left block, toward the right but at an upward angle of 30.0° with respect to the horizontal. It causes the left block to push on the right block. What are (a) the direction and (b) the magnitude of the force that the right block applies to the left block?

Potential answers for a:To the left at an angle 30 degrees down. To the right at an angle 30 degrees upwards. Directly to the left. Directly to the right.
The answer is directly to the left.

Answer for b: 71.1J

2. Relevant equations

Fnet = ma

3. The attempt at a solution
(I'm going to ignore the y-direction as it does not matter).

I made some free body diagrams and determined that the left block had forces Fcos(theta) - N acting acting on it. I set the acceleration equal to 0.

I said that the right block only had a positive N force acting on it. After setting the acceleration to 0, I realized that something was seriously wrong. How could the normal force be 0 in one equation but Fcos30 in the other

Thank you in advance for any help.

2. Oct 8, 2008

### Rake-MC

Since when does the y axis not matter?
a) Newton's third law states that every action has an equal reaction in the opposite direction. What will this tell us the direction of the force from mR on mL is?

b) Why did you set the acceleration to 0? This would mean that there is no force at all, this is where your problem arises. You know that the force on mR is the same as that on mL. Therefore you only need to calculate the force of one of the masses and omit the other. I can see you know how to do this.

3. Oct 9, 2008

### zoner7

Don't the forces in the y directions just cancel? On block R, N - mg = 0. On block L, Fsin(30) + N - mg = 0...

Following this logic, I wanted to conclude that we should then first only examine the x component of the force on block L. This force would push the two blocks together, causing them to exert normal forces on one another. I figured the the normal force of block R will push back on the block L in the opposite direction that the x-component of the force onto block L was applied. Since the x component is applied directly to the right, it stands to reason that the normal force of block R would push directly to its left.
by the way, I'm not saying that the external force and the normal force are equal and opposite pairs.

b)... I guess the net force will be greater than 0... hmm. makes sense. I'm so used to finding the magnitude of one force to counter another, resulting in constant velocity.

sounds good. I'll try the problem. Thank you

4. Oct 9, 2008

### Rake-MC

Yeah, you make decent points, but you still shouldn't ingore that dimension. I'm not sure, have you done the principle of transmissibility?
If so, the same line of force is acting through the entire block (rigid body). Therefore it is not simply a y force acting only on the left block and an x force acting through on the right block.
Consider this, you have two blocks touching each other with a lot of friction in between , by lifting up one block and pushing it towards the other block at the same time, the other block will lift up a little as a result won't it?

5. Oct 9, 2008

### DrDan

Your second post is getting closer. The y direction doesn't really matter. The component of the y direction force is about 56N, which is less than the weight of the block on the left. If there were friction, this y component force would reduce the force of friction, but the problem stated the surface is frictionless.

There is acceleration in the x direction. There is a net force on the two masses and no friction. So they accelerate as given by X component of applied force : (97N) = (ML+MR) a. Find a which is 3.86 m/s. Thats the acceleration of both blocks combined. To find the force from the block on the right take F(right) = MR (3.86m/s). That will get you your 71.1 Newtons.
Dr Dan

Last edited by a moderator: Oct 9, 2008
6. Oct 9, 2008

### zoner7

shouldn't the force that the right block applies to the left block be negative since it opposes the motion of the left block in the positive direction?

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