anonymous
  • anonymous
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
Physics
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

anonymous
  • anonymous
what is the exact formula for letter a and b ??
anonymous
  • 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.
anonymous
  • 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.)

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
Let me rephrase. The boy's x-component force imparted on the box must equal said value.
anonymous
  • anonymous
aw. this is the only problem left here in my problem set. i cant answer it because i forgot the formula
anonymous
  • anonymous
JamesJ help me with your knowledge of typing equations on this site, lol.
JamesJ
  • 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
anonymous
  • anonymous
ok, then ?
JamesJ
  • 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.
anonymous
  • anonymous
keep going. i think i might get it
JamesJ
  • JamesJ
That will give you the answer to part a.
JamesJ
  • 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.
anonymous
  • anonymous
The y component of the boy's pull.
JamesJ
  • JamesJ
\[ N_{modified} = N_{unmodified} - F\sin(30) \]
anonymous
  • anonymous
Yes, that.
JamesJ
  • JamesJ
Hence \[ F_f = -\mu_{static}N_{modified} = -\mu_{static}(mg - F\sin(30)) \] Now solve for F.
anonymous
  • anonymous
ok thank you. i get it now.

Looking for something else?

Not the answer you are looking for? Search for more explanations.