Given: 2 wedge shaped blocks arranged as a rectangle.

The bottom block is on a frictionless horizontal surface.

Horizontal force Q is applied to the vertical side of the bottom block.

Find Q max, the maximum value of Q where there is no slipping, in terms of us, m, and theta.

Masses: m and m

us is the coefficient of static friction between the blocks.

Part 2: If the force is applied to the other block horizontally, what is Q max?

For the first part, I applied newton's laws to the top block and got the correct answer.

For the second part, If I apply newton's laws to the top block again, I get the correct (same) answer; BUT only if I ignore Q in my equations for F=ma.

From my reasoning, the top block will be the one slipping under both circumstances, so that's the block that I addressed...

Is this strategy correct? If so, why can I ignore Q? If not, how do I approach this?

-----------

Here is my thinking, please tell me if this is correct:

Since we need Fmax, we can find the maximum acceleration such that there is about to be slipping. Because there is no slipping at this point, the accelerations of each block, as well as of the system, are all equal. So, by figuring out the maximum acceleration of the top block, which is the one that will slip upward on the other block, we can use this to solve the problem.

-----------

Also, does anyone know where I can find more problems like this?

The bottom block is on a frictionless horizontal surface.

Horizontal force Q is applied to the vertical side of the bottom block.

Find Q max, the maximum value of Q where there is no slipping, in terms of us, m, and theta.

Masses: m and m

us is the coefficient of static friction between the blocks.

Code:

```
----
| /|
| / |<---Q
|/__|
```

Code:

```
----
| /|
Q-->| / |
|/__|
```

For the first part, I applied newton's laws to the top block and got the correct answer.

For the second part, If I apply newton's laws to the top block again, I get the correct (same) answer; BUT only if I ignore Q in my equations for F=ma.

From my reasoning, the top block will be the one slipping under both circumstances, so that's the block that I addressed...

Is this strategy correct? If so, why can I ignore Q? If not, how do I approach this?

-----------

Here is my thinking, please tell me if this is correct:

Since we need Fmax, we can find the maximum acceleration such that there is about to be slipping. Because there is no slipping at this point, the accelerations of each block, as well as of the system, are all equal. So, by figuring out the maximum acceleration of the top block, which is the one that will slip upward on the other block, we can use this to solve the problem.

-----------

Also, does anyone know where I can find more problems like this?

Last edited: