anonymous 4 years ago Physics Problem: A boy pulls a 30.0kg box on a horizontal surface with a force that makes an angle 30degrees the horizontal. If the coefficient of static friction and kinetic friction between the box and the surface are 0.45 and 0.35 respectively, find: a.) the force that will start the box to move b.) the acceleration of the box of the boy maintains the same force that moves the box

1. anonymous

what is the exact formula for letter a and b ??

2. anonymous

The maximum value for the opposing force given by the coefficient of static friction is the product of that and the box's normal force.

3. anonymous

This means the boy's force must at least equal said value. (Sorry, I can't write formulas well since I'm not good with LaTeX.)

4. anonymous

Let me rephrase. The boy's x-component force imparted on the box must equal said value.

5. anonymous

aw. this is the only problem left here in my problem set. i cant answer it because i forgot the formula

6. anonymous

JamesJ help me with your knowledge of typing equations on this site, lol.

7. JamesJ

First step draw a diagram. In that diagram, you can see that the x-component of the force F on the box is $F_x = F \cos(30) = \frac{\sqrt{3}}{2}F$ Now, the friction on the box at exactly the force level that makes it move is $F_f = -\mu_{static}N = -\mu_{static}mg$ This force is - negative, because it acts in the opposite direction to the force pulling the box F_x - and N = mg, because this is the normal reaction force to the weight of the box

8. anonymous

ok, then ?

9. JamesJ

When the box starts to move, the force F is just sufficiently large that $F_x + F_f$ is infinitesimally larger than 0. Or for practical purposes equal to zero. Hence set this quantity equal to zero. You know every variable in that equation except F. Solve for that variable.

10. anonymous

keep going. i think i might get it

11. JamesJ

That will give you the answer to part a.

12. JamesJ

Actually, one twist on this. You need to modify N. N is not exactly equal to mg. It would be if there was no other vertical force on the box. But there is.

13. anonymous

The y component of the boy's pull.

14. JamesJ

$N_{modified} = N_{unmodified} - F\sin(30)$

15. anonymous

Yes, that.

16. JamesJ

Hence $F_f = -\mu_{static}N_{modified} = -\mu_{static}(mg - F\sin(30))$ Now solve for F.

17. anonymous

ok thank you. i get it now.