- #1

NIHLUS13

- 5

- 0

## 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.

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.