Answer:
u_critical = 1.33
Explanation:
Given:
- The complete question is:
[Box on Ramp] A 6.00-kg box sits on a ramp that is inclined at 37.0° above the horizontal. The coefficient of kinetic friction between the box and the ramp is th=0.300.
Find:
Show that if the coefficient of friction were larger than a certain value, then the box would not start moving no matter how large a horizontal force is applied. What is this critical value for the friction coefficient? Please be sure your reasoning is clear.
Solution:
- The equation of motion for the block on the ramp is given by:
Fhcos(θ) - mgsin(θ) - Ff = m*a
Where,
m = mass of block , Ff = Frictional Force , Fh = Horizontal applied force, θ = Angle of the slope , a = acceleration of box
- When the block does not move then a = 0, we have:
Ff = Fhcos(θ) - mgsin(θ)
- The frictional force of the box Ff is given by:
Ff =< u*N
Where,
N = Normal contact force, u = coefficient of static friction.
- The contact force N is given by the equilibrium equation in the direction normal to slope is:
Fhsin(θ) + mgcos(θ) = N
- The frictional force Ff is given by:
Ff =< [ Fhsin(θ) + mgcos(θ) ]*u
- Then substitute the two bold equations:
Fhcos(θ) - mgsin(θ) =< [ Fhsin(θ) + mgcos(θ) ]*u
u >= [ Fhcos(θ) - mgsin(θ) ] / [ Fhsin(θ) + mgcos(θ) ]
- The critical value for u is given for limit Fh -> ∞ is:
u>= Lim ( Fh -> ∞) { [ Fhcos(θ) - mgsin(θ) ] / [ Fhsin(θ) + mgcos(θ) ]}
u>= Lim ( Fh -> ∞) { [cos(θ) - mgsin(θ) / Fh ] / [sin(θ) + mgcos(θ)/Fh ]}
- Evaluate limit, we get:
u >= cos(θ) / sin(θ)
u >= cos(37) / sin(37)
u >= 1.33
- The critical value for coefficient of friction is u_critical = 1.33.
