What is the necessary force to move a crate on a rough surface?

[miglo]

Homework Statement

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.

Homework Equations

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

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 I am using the wrong angle, should it be 21? or 360-21=339?
or did i not solve for $F$ correctly?

 

[1MileCrash]

Everything looks fine.

What is your final answer?

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

 

[miglo]

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

 

[1MileCrash]

Don't slap yourself, but








Your calculator is in radian mode.

 

[miglo]

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

 

[1MileCrash]

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