Find coefficient of kinetic friction of block and ramp.

  • #1
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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.
 

Answers and Replies

  • #2
TSny
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Welcome to PF!

Could the setup for part (b) be as shown:
upload_2016-12-30_14-53-56.png

Was any information given about the initial height of the block on the ramp?
 
  • #3
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Welcome to PF!

Could the setup for part (b) be as shown:
View attachment 110936
Was any information give about the initial height of the block on the ramp?
Perhaps but no there is no info regarding the height of the block on the ramp if it is that set up.
 
  • #4
TSny
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In part (b) it says that the block is given a small nudge to get started. This means that the block is initially on the verge of slipping as in part (a). So, I'm "inclined" to think that the ramp is still sloping at 35 degrees in part (b).

Anyway, I don't think enough information is given to work the problem if the ramp is on top of the table as shown in the figure I posted.
 
  • #5
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In part (b) it says that the block is given a small nudge to get started. This means that the block is initially on the verge of slipping as in part (a). So, I'm "inclined" to think that the ramp is still sloping at 35 degrees in part (b).

Anyway, I don't think enough information is given to work the problem if the ramp is on top of the table as shown in the figure I posted.
I agree, there isnt enough info for that figure. Also the reason i assumed the angle changes is due to the fact that if theta is 35 degrees length x and height y do not correspond to the correct values.
 
  • #6
TSny
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I agree, there isnt enough info for that figure. Also the reason i assumed the angle changes is due to the fact that if theta is 35 degrees length x and height y do not correspond to the correct values.
Too bad they didn't provide a figure. If the setup is as I indicated, then working backward from the given answer, I get that the block must have been placed at a vertical height of 1.0 m from the base of the ramp (i.e., 1.0 m above the table top).
 
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  • #7
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Too bad they didn't provide a figure. If the setup is as I indicated, then working backward from the given answer, I get that the block must have been placed at a vertical height of 1.0 m from the base of the ramp.
And what was your approach?
 
  • #8
TSny
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How would you approach it?
 
  • #9
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How would you approach it?
As before using work energy theorem, however what of the 0.8 m height of the table itself?
 
  • #10
TSny
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What can you learn about the motion of the block at the bottom of the ramp from the fact that the table is 0.8 m high and the block lands 0.4 m horizontally from the edge of the table?
 
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