Dynamics and Friction: Solving for Maximum Angle on a Ramp with μ_{s}=0.25

[hsphysics2]

Homework Statement

A box is halfway up a ramp. The ramp makes an angle, θ with the ground. What is the maximum value of θ before the mass will slip? μ$_{s}$=0.25

Homework Equations

F$_{x}$=ma$_{x}$

The Attempt at a Solution

I drew a free body diagram to show the forces affecting the box

η-mgcos=0
η=mgcosθ (eq'n 1)


F$_{x}$=ma$_{x}$
μ$_{s}$η-mgsinθ=ma$_{x}$ (sub eq'n 1 in)
μ$_{s}$(mgcosθ)-mgsinθ=ma$_{x}$
mg(μ$_{s}$cosθ-sinθ)=ma$_{x}$


I'm not sure where to go from here, or even if this is the correct path for me to take

 

[PhanthomJay]

You have correctly identified the forces acting, but just as the box is on the verge of slipping, is it accelerating?

 

[hsphysics2]

the box wouldn't be accelerating, so would it be

mg(μ$_{s}$cosθ-sinθ)=0
mgμ$_{s}$cosθ=mgsinθ
μ$_{s}$cosθ=sinθ
μ$_{s}$=tanθ
θ=14.036

 

[PhanthomJay]

the box wouldn't be accelerating, so would it be

mg(μ$_{s}$cosθ-sinθ)=0
mgμ$_{s}$cosθ=mgsinθ
μ$_{s}$cosθ=sinθ
μ$_{s}$=tanθ
θ=14.036

Yes, good, in degrees (don't forget units), but you should round your answer to 14 degrees ( 2 significant figures).

 

[Nickie]

. Can anyone point me in the right direction?



Your approach is correct so far. To solve for the maximum angle, we can use the condition for static equilibrium, where the net force in the x-direction is equal to zero. This means that the acceleration in the x-direction is also equal to zero. Therefore, we can set the right side of your last equation to zero and solve for the maximum angle:

mg(μ_{s}cosθ-sinθ)=0

μ_{s}cosθ-sinθ=0

μ_{s}cosθ=sinθ

tanθ=μ_{s}

θ=arctan(μ_{s})

Plugging in the given value for μ_{s}=0.25, we get a maximum angle of approximately 14.04 degrees. This means that the box can safely be placed on a ramp with an angle less than or equal to 14.04 degrees without slipping, given the coefficient of static friction of 0.25.