# ME Statics - Wedge problem

1. Nov 19, 2014

### Feodalherren

1. The problem statement, all variables and given/known data
The crate shown is held against wedge B by a spring. The spring is 96.0% of its original uncompressed lengthl=2.75m, and the spring constant is given ask=1650N/m. The coefficient of static friction at all contacting surfaces is μs=0.150. The mass of the crate is m=22.0kg . The angle is θ=11.0∘. Neglect the mass of the wedge. Assume the crate only moves in the y direction and that wedge A cannot move.

Determine the magnitude of the smallest horizontal force P that is necessary to begin moving the crate upward.
2. Relevant equations

3. The attempt at a solution
I found the normal from the spring and the weight of the box to be 397N (this is known to be the correct value).

Since A can't move, we can draw a FBD of block B.

I noticed that I drew the line for the 79 degree angle for f wrong. ignore it. The 79 degrees is supposed to be along the Y-axis. The number is still correct.

Sum of the forces in X = 0 = -P + Ncos79 + .15Nsin79

Sum of the forces in Y=0=-397+NSin79 - .15Ncos79

Using a matrix to solve for P we get P=141 N which is incorrect. I've ran out of ideas here. Every way I do it I seem to get the same answer but it supposedly incorrect.

2. Nov 20, 2014

### Staff: Mentor

You have substituted your integer answer into the equations and confirmed it's a perfect fit?

Do you know what the correct answer is supposed to be?

I make the spring & weight combo to be 397.1N (using g=9.8 m/s²).

Last edited: Nov 20, 2014
3. Nov 20, 2014

### Feodalherren

Yes. It's close enough. The answer doesn't need to be exact. It's 3 sig-figs and within 5 or 10 % and the program will let you know the answer.

4. Nov 20, 2014

### Staff: Mentor

My working agrees with your pair of ∑ f equations.

5. Nov 20, 2014

### Feodalherren

Very strange... I will ask my professor tomorrow. Thank you, Sir.

6. Nov 15, 2016

### singingbookworm

You forgot the force of static friction between the crate and block B, opposing block B moving to the left (it points to the right at the top of the block). When you sum forces in the x this will change your answer slightly for F which actually changes a lot in your answer. Hope this helps anyone else who is looking to this for help!