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NIHLUS13
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Homework Statement
A small block is on the point of slipping down a rough ramp inclined at 35 degrees to the horizontal.
a) What is the coefficient of static friction between the block and the ramp?
Answer μ_s = 0.70
b) The ramp is attached to the edge of a 0.8 m high table. After the block is given a small nudge it slides down the ramp and lands on the floor a horizontal distance of 0.4 m away from the edge of the table. What is the coefficient of kinetic friction between the block and the ramp?
Answer μ_k = 0.62
Homework Equations
For part a)
∑ F_parallel to ramp = 0 N
∑ F _perpendicular to ramp = 0 N
Static friction force F_f = μ_s* R
For part b)
∑ F_parallel to ramp = ma
∑ F _perpendicular to ramp = 0 N
Work energy theorem
W_exerted = ΔE_mech +ΔE_friction = 0 J
ΔE_mech = ΔKE + ΔU
ΔE_friction = μ_k* R *d
The Attempt at a Solution
For part a)
∑ F_parallel to ramp = 0 N
Therefore if the direction pointing down the ramp is positive, and perpendicular to ramp is positive. Then...
0 = mg sin (35) - μ_s* R
∑ F _perpendicular to ramp = 0 N
0 = R - mg cos(35)
Hence mg sin (35) = μ_s*mg cos (35)
Finally μ_s = tan (35) = 0.70
Which is the correct answer, the major issue occurs within part b).
For part b)
Note the following leads to a line such that μ_k will cancel out! Hence another approach to
achieving the solution μ_k = 0.62 must be used.
I started by assuming the following;
Vertical height of the block initially is y = 0.8 m, and when the block reaches the floor y = 0 m.
Horizontal distance between edge of the table and the other end of ramp is x = 0.4 m.
Ramp length is d = √ ( 0.8^2 + 0.4^2 ) = (2√5)/5 m.
Finally incline is not 35 degrees but in fact arctan (0.8/0.4) ≅ 63.4 degrees.
Using W_exerted = ΔE_mech +ΔE_friction = 0 J
ΔKE + ΔU + μ_k* R *d = 0 J
0.5 m v_f^2 - 0 + 0 - mgy + μ_k* R *d = 0 J note v_f is velocity final
0.5 m v_f^2 - mgy + μ_k* R *d = 0 J
∑ F _perpendicular to ramp = 0 N
R - mg cos (63.4) = 0 N
Therefore
0.5 m v_f^2 - mgy + μ_k* mg cos (63.4) *d = 0 J , d = (2√5)/5 m
0.5 m v_f^2 - mgy + μ_k* mg cos (63.4) *(2√5)/5 = 0 J
Eliminating mass
0.5 v_f^2 - gy + μ_k* g cos (63.4) *(2√5)/5 = 0 J
0.5 v_f^2 = gy - μ_k* g cos (63.4) *(2√5)/5
v_f^2 = 2gy - 2μ_k* g cos (63.4) *(2√5)/5
Replacing v_f^2 with u^2 + 2ad, where u^2 = 0 as initial velocity is zero parallel to ramp, and d = (2√5)/5 m
2a(2√5)/5 = 2gy - 2μ_k* g cos (63.4) *(2√5)/5
a(2√5)/5 = gy - μ_k* g cos (63.4) *(2√5)/5
Using ∑ F_parallel to ramp = ma to substitute for a
ma = mg sin (63.4) - μ_k* mg cos (63.4)
a = g sin (63.4) - μ_k* g cos (63.4)
a = g (sin (63.4) - μ_k* cos (63.4) )
g (sin (63.4) - μ_k* cos (63.4) ) * (2√5)/5 = gy - μ_k* g cos (63.4) *(2√5)/5
(sin (63.4) - μ_k* cos (63.4) ) * (2√5)/5 = y - μ_k* cos (63.4) *(2√5)/5 The line above μ_k cancels.
