# Force with static friction

1. Oct 10, 2011

### miglo

1. The problem statement, all variables and given/known data
To move a large crate across a rough floor, you push on it with a force F at an angle of 21° below the horizontal, as shown in the figure. Find the force necessary to start the crate moving, given that the mass of the crate is m = 37 kg and the coefficient of static friction between the crate and the floor is 0.60.

2. Relevant equations
$$F_x=F\cos{\theta}$$
$$F_y=F\sin{\theta}$$
$$F=ma$$
$$W=mg$$
$$f_s=\mu_sN$$

3. The attempt at a solution
so i know that $$\sum{\vec{F_x}}=F\cos{\theta}-\mu_sN=0$$
and $$\sum{\vec{F_y}}=N-F\sin{\theta}-mg=0$$
solving for $N$ i get $N=F\sin{\theta}+mg$ and plugging this in for N in the sum of the forces in the x direction i get $F\cos{\theta}-\mu_s(F\sin{\theta}+mg)=0$ and finally solving for $F$ i get $F=\frac{\mu_smg}{\cos{\theta}-\mu_s\sin{\theta}}$
so i plugged in all the values given and yet i still got the wrong answer (my homework is online and i get 5 tries for each question)
my guess is that im using the wrong angle, should it be 21? or 360-21=339?
or did i not solve for $F$ correctly?

Last edited: Oct 10, 2011
2. Oct 10, 2011

### 1MileCrash

Everything looks fine.

No reason to use the other angle as long as you realize the direction the component will be applied in (which you do.)

3. Oct 11, 2011

### miglo

i get 207.47N as my answer, but it says my answer differs from the correct answer by more than 10%
any help?

4. Oct 11, 2011

### 1MileCrash

Don't slap yourself, but

5. Oct 11, 2011

### miglo

hahaha wow i forgot i had it in radians from my calculus 2 class this morning
thanks a lot 1MileCrash

6. Oct 11, 2011

### 1MileCrash

You're welcome, it happens to everyone. The important thing is that you do know how to work the problem.