## alfers101 Group Title 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 2 years ago 2 years ago

1. alfers101 Group Title

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

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.

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.)

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

5. alfers101 Group Title

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

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

7. JamesJ Group Title

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. alfers101 Group Title

ok, then ?

9. JamesJ Group Title

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. alfers101 Group Title

keep going. i think i might get it

11. JamesJ Group Title

That will give you the answer to part a.

12. JamesJ Group Title

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.

The y component of the boy's pull.

14. JamesJ Group Title

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

Yes, that.

16. JamesJ Group Title

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

17. alfers101 Group Title

ok thank you. i get it now.