a 55.0 kg cheerleader uses an oil-filled hydraulic lift to hold four 130 kg football players at a height of 0.800 m . the diamet
1 answer:
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Refer to the free body diagram shown melow.
F = applied force
R = frictional force
m = 15 kg, the mass of the object
The acceleration (to the right) is 16.3 m/s², therefore
F - R = (15 kg)*(16.3 m/s²) = 244.5 N
The normal reaction is
N = mg = (15 kg)*(9.8 m/s²) = 147 N
The frictional force is
R = μN = 147μ N, where μ = coefficient of kinetic friction.
Let us check possible answers:
If R = 5.5 N, then μ = 5.5/147 = 0.0374 (very likely)
If R = 15 N, then μ = 15/147 = 0.102 (possible)
If R = 244.5 N, (Highly unlikely, exceed mg)
If R = 494.5 N, (highly unlikely, exceeds mg)
Answer:
The most reasonable answer is R = 5.5 N
F = applied force
R = frictional force
m = 15 kg, the mass of the object
The acceleration (to the right) is 16.3 m/s², therefore
F - R = (15 kg)*(16.3 m/s²) = 244.5 N
The normal reaction is
N = mg = (15 kg)*(9.8 m/s²) = 147 N
The frictional force is
R = μN = 147μ N, where μ = coefficient of kinetic friction.
Let us check possible answers:
If R = 5.5 N, then μ = 5.5/147 = 0.0374 (very likely)
If R = 15 N, then μ = 15/147 = 0.102 (possible)
If R = 244.5 N, (Highly unlikely, exceed mg)
If R = 494.5 N, (highly unlikely, exceeds mg)
Answer:
The most reasonable answer is R = 5.5 N
Solution
Force between pair of objects of masses 1kg and 2kg that are 1m apart is given as
here G is gravitational constant
G=
therefore,
similarly Force between pair of objects of masses 2kg each that are 1m apart is given as
or F'=2F
it means Force between pair of objects of masses 2kg each that are 1m apart is equal to twice the Force between pair of objects of masses 1kg and 2kg that are 1m