# Figuring out the force using Newton's 2nd Law

Here's a problem on my test review:

A 12 kg crate resta on a horizontal surface and a boy pulls on it with a force that is 30° below the horizontal. If the coefficient of static friction is 0.40, the minimum magnitude force he needs to start the crate moving is:

Okay, so I found the equation for the forces in the Y direction: $$\sum F_{Y} = F_{N} - mg - F\sin\theta = 0$$ $$F_{N} = mg + F\sin\theta$$

And the X direction: $$\sum F_{X} = F\cos(\theta) - F_{f} = 0$$ $$F\cos\theta = F_{f}$$

Solving for the force: $$F\cos\theta = u_{s}F_N$$ $$F\cos\theta = u_s (mg + F\sin\theta)$$ $$F\cos{-30°} = 0.40 \left[(12 kg)(9.8 \frac{m}{s^2}) + F\sin{-30°} \right]$$ $$F \left[\cos{-30°} - (.4)(\sin{-30°} \right] = 47.04 N$$ $$F=44N$$

However, it looks like the correct answer was 71. Any ideas where I went wrong here?

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The expression for $F_Y$ should be $$\displaystyle\sum F_Y=F_N-mg+F \sin\theta\;.$$ All terms in such expressions should start out positive. They can turn negative later depending on the angle at which they're applied. In fact, if you wanted to be super-explicit you could write $$\displaystyle\sum F_Y=F_N\sin(90°)+mg\sin(-90°)+F \sin\theta\;.$$ It's hardly necessary in this case, though.
Working from $F\cos(-30°)=\mu_sF_N=0.4\left[9.807\times 12-F\sin(-30°)\right]$ gives $F=70.7$. Check your math. – rdhs Dec 5 '11 at 5:40