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alfers101
 3 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
alfers101
 3 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

This Question is Closed

alfers101
 3 years ago
Best ResponseYou've already chosen the best response.0what is the exact formula for letter a and b ??

badreferences
 3 years ago
Best ResponseYou've already chosen the best response.0The maximum value for the opposing force given by the coefficient of static friction is the product of that and the box's normal force.

badreferences
 3 years ago
Best ResponseYou've already chosen the best response.0This 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.)

badreferences
 3 years ago
Best ResponseYou've already chosen the best response.0Let me rephrase. The boy's xcomponent force imparted on the box must equal said value.

alfers101
 3 years ago
Best ResponseYou've already chosen the best response.0aw. this is the only problem left here in my problem set. i cant answer it because i forgot the formula

badreferences
 3 years ago
Best ResponseYou've already chosen the best response.0JamesJ help me with your knowledge of typing equations on this site, lol.

JamesJ
 3 years ago
Best ResponseYou've already chosen the best response.3First step draw a diagram. In that diagram, you can see that the xcomponent 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

JamesJ
 3 years ago
Best ResponseYou've already chosen the best response.3When 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.

alfers101
 3 years ago
Best ResponseYou've already chosen the best response.0keep going. i think i might get it

JamesJ
 3 years ago
Best ResponseYou've already chosen the best response.3That will give you the answer to part a.

JamesJ
 3 years ago
Best ResponseYou've already chosen the best response.3Actually, 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.

badreferences
 3 years ago
Best ResponseYou've already chosen the best response.0The y component of the boy's pull.

JamesJ
 3 years ago
Best ResponseYou've already chosen the best response.3\[ N_{modified} = N_{unmodified}  F\sin(30) \]

JamesJ
 3 years ago
Best ResponseYou've already chosen the best response.3Hence \[ F_f = \mu_{static}N_{modified} = \mu_{static}(mg  F\sin(30)) \] Now solve for F.

alfers101
 3 years ago
Best ResponseYou've already chosen the best response.0ok thank you. i get it now.
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