Salina’s internal desire to exercise is called
2 answers:
You might be interested in
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
