Answer:
Explanation:
From the question we are told that:
Length
Mass
Distance from Tip
Generally, the equation for Torque Balance is mathematically given by
Therefore
A slide inclined at 35 takes bathers into a swimming pool. with water sprayed on the slide the bathers spends only one third as
much timeo n the slide as whne its dry. find the coefficient of friction on the dry slide
1 answer:
Refer to the figure shown below.
mg = the weight of the person on the slide.
N = mg cos(35°) = 0.8192 mg, the normal reaction
F = mg sin(35°) = 0.5736 mg, the force acting down the slide.
R = μN = 0.8192 μmg, the resisting force due to friction
μ = the coefficient of dynamic friction on the dry slide.
d = the length of the slide.
g = 9.8 m/s²
All measurements are in SI units.
We shall consider two cases.
Case 1 (The slide is lubricated by water spray).
Assume that the coefficient of friction is approximately zero.
Let a = the acceleration down the slide. Then
0.5736 mg = ma
a = 0.5736g = 5.6213 m/s²
Let t = the time to travel the length of the slide from rest.
d = (1/2)*(5.6213 m/s²)(t s)² = 2.8106t² m (1)
t² = 0.3558d
t = 0.5965√d s (2)
Case 2 (The slide is dry)
The time to travel down the slide is 3*(0.5965√t) = 1.7895√t s.
Let a = the acceleration. Then
0.5796mg - 0.8192μmg = ma
a = 0.5796g - 0.8192μg = 5.6801 - 8.0282μ m/s²
The travel doen the slide from rest is given by
d = (1/2)*(5.6801 - 8.0282μ)*(1.7895√t)²
d = (9.0947 - 12.8544μ)t² (3)
Equate (1) and (3).
9.0947 - 12.8544μ = 2.8106
12.8544μ = 6.2841
μ = 0.489
Answer: μ = 0.49 (nearest hundredth)
mg = the weight of the person on the slide.
N = mg cos(35°) = 0.8192 mg, the normal reaction
F = mg sin(35°) = 0.5736 mg, the force acting down the slide.
R = μN = 0.8192 μmg, the resisting force due to friction
μ = the coefficient of dynamic friction on the dry slide.
d = the length of the slide.
g = 9.8 m/s²
All measurements are in SI units.
We shall consider two cases.
Case 1 (The slide is lubricated by water spray).
Assume that the coefficient of friction is approximately zero.
Let a = the acceleration down the slide. Then
0.5736 mg = ma
a = 0.5736g = 5.6213 m/s²
Let t = the time to travel the length of the slide from rest.
d = (1/2)*(5.6213 m/s²)(t s)² = 2.8106t² m (1)
t² = 0.3558d
t = 0.5965√d s (2)
Case 2 (The slide is dry)
The time to travel down the slide is 3*(0.5965√t) = 1.7895√t s.
Let a = the acceleration. Then
0.5796mg - 0.8192μmg = ma
a = 0.5796g - 0.8192μg = 5.6801 - 8.0282μ m/s²
The travel doen the slide from rest is given by
d = (1/2)*(5.6801 - 8.0282μ)*(1.7895√t)²
d = (9.0947 - 12.8544μ)t² (3)
Equate (1) and (3).
9.0947 - 12.8544μ = 2.8106
12.8544μ = 6.2841
μ = 0.489
Answer: μ = 0.49 (nearest hundredth)
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