- #1

slybuster

- 10

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## Homework Statement

A box with a mass of 22 kg is at rest on a ramp inclined at 45 degrees to the horizontal. The coefficients of friction between the box and the ramp are μs = 0.78 and μk = 0.65

a) Determine magnitude of the largest force that can be applied upward, parallel to the ramp, if the box is to remain at rest.

b) Determine the magnitude of the smallest force that can be applied onto the top of the box, perpendicular to the ramp, if the box is to remain at rest.

## Homework Equations

ΣF = Fμs (or Fμk) +/- Fax (or Fay)

ΣF = 0

## The Attempt at a Solution

When I diagram the problem, it seems to me that the block should be in motion (as the horizontal component of gravity [Fgx = Fg*sin(45)]is greater than the static friction inherent to the system [Ffμs = Fn*μs]), but the problem states the the system is at rest.

The following are my attempts to answer the question, can you please tell me if I am misunderstanding static & kinetic friction and getting the wrong answer?

a)

m = 22 kg

θ = 45

μs = .78

g = 9.8 m/s^2

***Fax = ?

Fa = Fgx + Fμs

Fa = (mg*sinθ) + (mg*cosθ*μs)

Fa = 152.45N + 118.91N

Fa = 2.7*10^2N

b)

For this part of the problem, I tried to include μk (i.e. confusing because the problem states the block starts from rest?!)

m = 22 kg

θ = 45

μs = .78

μk = .65

g = 9.8 m/s^2

***Fay = ?

Fx = μk*mg*sinθ

Fx = 99.09N (Down)

μs(Fgy + Fay) = Fx

Fay(0.78) = 19.5N

Fay = 2.5*10N

I feel like this is totally wrong because both kinetic and static friction are involved in the problem...can someone please look at my answers and see if I'm on the right track and have the correct answers? Thank you in advance for reading this...