Original Problem: A 225Kg crate rests on a surface that is inclined above thehor

Question

Original Problem:
A 225Kg crate rests on a surface that is inclined above thehorizontal at an angle of 20.0o. A horizontal force(magnitude = 535N and parallel to the ground, not the incline) isrequired to start the crate moving down the incline. What is thecoefficient of static friction between the crate and theincline?

I am having problems getting a valid coefficient of static frictionfor Alternate Problem 118. I found the normal force of the crate byusing F=ma*sin where is the angle of the inclineplane. I then found the weight of the crate by using F=ma. Iunderstand that in order to find the coefficient of staticfriction, I need to know the force of friction and apply that toFf = s*FN. However, theproblem states that there is a horizontal force applied to thecrate to overcome static friction. I am unsure of how to translatethe horizontal force to be in the same direction as the inclineplane so that I can find the force of friction. Please let me knowhow I can find the force of static friction. Thanks much for yourhelp.

Cameron T

Explanation / Answer

          Giventhat the mass of crate is m = 225 kg           Angle ofinclination is = 20o         Horizontalforce is F = 535 N ------------------------------------------------------------------    Apply Netons law in verticle direction                              N = mg cos - F sin Apply Newtons law along length of ramp                         F cos + mg sin - f = 0 ( At equllibriumposition)                          Fcos + mg sin - *N = 0                         F cos + mg sin = *N                                                   = F cos + mg sin /N                                                        = ( F cos + mg sin ) / (mg cos - Fsin )                                                     =------------            Horizontalforce is F = 535 N ------------------------------------------------------------------    Apply Netons law in verticle direction                              N = mg cos - F sin Apply Newtons law along length of ramp                         F cos + mg sin - f = 0 ( At equllibriumposition)                          Fcos + mg sin - *N = 0                         F cos + mg sin = *N                                                   = F cos + mg sin /N                                                        = ( F cos + mg sin ) / (mg cos - Fsin )                                                     =------------    Apply Netons law in verticle direction                              N = mg cos - F sin Apply Newtons law along length of ramp                         F cos + mg sin - f = 0 ( At equllibriumposition)                          Fcos + mg sin - *N = 0                         F cos + mg sin = *N                                                   = F cos + mg sin /N                                                        = ( F cos + mg sin ) / (mg cos - Fsin )                                                     =------------   

Related Questions

Navigate

• Browse (All)
• Browse O
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.