A hotel wants to put a fence around a circular spa. The radius of the spa is 5/√2-1 feet. If the fence is built along the immedi
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Refer to the figure shown below.
g = 9.8 m/s², the acceleration due to gravity.
W = mg, the weight of the mug.
θ = 17°, the angle of the ramp.
Let μ = the coefficient of static friction.
The force acting down the ramp is
F = W sin θ = W sin(17°) = 0.2924W N
The normal reaction is
N = W cosθ = W cos(17°) = 0.9563W N
The resistive force due to friction is
R = μN = 0.9563μW N
For static equilibrium,
μN = F
0.9563μW =0.2924W
μ = 0.3058
The frictional force is F = μN = 0.2924W
The minimum value of μ required to prevent the mug from sliding satisfies
the condition
R > F
0.9563μW > 0.2924W
μ > 002924/.9563 = 0.306
Answer:
The frictional force is 0.2924mg, where m = the mass of the mug.
The minimum coefficient of static friction is 0.306
g = 9.8 m/s², the acceleration due to gravity.
W = mg, the weight of the mug.
θ = 17°, the angle of the ramp.
Let μ = the coefficient of static friction.
The force acting down the ramp is
F = W sin θ = W sin(17°) = 0.2924W N
The normal reaction is
N = W cosθ = W cos(17°) = 0.9563W N
The resistive force due to friction is
R = μN = 0.9563μW N
For static equilibrium,
μN = F
0.9563μW =0.2924W
μ = 0.3058
The frictional force is F = μN = 0.2924W
The minimum value of μ required to prevent the mug from sliding satisfies
the condition
R > F
0.9563μW > 0.2924W
μ > 002924/.9563 = 0.306
Answer:
The frictional force is 0.2924mg, where m = the mass of the mug.
The minimum coefficient of static friction is 0.306
The initial force between the two charges is given by:
where k is the Coulomb's constant, q1 and q2 the two charges, d their separation. Let's analyze now the other situations:
1. F
In this case, q1 is halved, q2 is doubled, but the distance between the charges remains d.
So, we have:
So, the new force is:
So the force has not changed.
2. F/4
In this case, q1 and q2 are unchanged. The distance between the charges is doubled to 2d.
So, we have:
So, the new force is:
So the force has decreased by a factor 4.
3. 6F
In this case, q1 is doubled and q2 is tripled. The distance between the charges remains d.
So, we have:
So, the new force is:
So the force has increased by a factor 6.