A horizontal force of 400.0 N is required to pull a 1760 N trunk across the floor at a constant speed. Find the coefficient of sliding friction.
: given Fg = 1760N
Fpull = 400.0N
Since the trunk is moving at CONSTANT speed, the horizontal forces must be in balance (Fnet = 0). That means that the force of friction has the same magnitude as the pulling force. Fpull = Ffriction = 400.0 N.
Also, since the trunk is not moving vertically, the force of gravity on the trunk must be equal in magnitude to the normal force the floor is pushing up on the trunk.
needed equations : Ff = µFN
Fg = FN = 1760 N
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FN = 1760N
F f = 400.0N
F f = µFN
µ=
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Ff FN
→ →
µ=
Ff FN
µ=
400.0N 1760N
→
µ = 0.227
How much force must be applied to push a 1.35 kg book across the desk at constant speed if the coefficient of sliding friction is 0.30?
Fpush = Ffriction
needed equations : F = ma → Fg = mg F f = µFN
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Since the book is moving at CONSTANT speed, the horizontal forces must be in balance (Fnet = 0). That means that the force of friction has the same magnitude as the pushing force.
given : m = 1.35kg µ = 0.30
Also, since the book is not moving vertically, the force of gravity on the book must be equal in magnitude to the normal force the floor is pushing up on the book. Fg = FN
Fg = mg → Fg = (1.35kg)(9.8 m s2 ) → Fg = 13.23 N
→ FN = 13.23 N
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F f = µFN
→
F f = (0.3)(13.23N )
F f = Fpush
→
Fpush = 4.0 N
→
F f = 4.0 N
If a 1500.0 N force is exerted on a 200.0 kg crate to move it across the floor and the coefficient of friction is 0.250, what is the crate’s acceleration?
given : m = 200.0 kg µ = 0.250 Fpush = 1500.0 N
Since the crate is moving at accelerating, the horizontal forces are not balanced (Fnet ≠ 0). That means that the force of pushing the crate has a larger magnitude than the friction force. Fnet = Fpush - Ffriction *note, friction is negative in magnitude because it always acts opposite motion!
needed equations :
F = ma → Fg = mg
In this case we want to find Fnet so we can use it to calculate the acceleration of the crate.
F f = µFN
Also, just as in the above problems since the crate is not moving vertically, the force of gravity on the crate must be equal in magnitude to the normal force the floor is pushing up on the crate.
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Fg = FN
Fg = mg → Fg = (200.0kg)(9.8 m s2 ) → Fg = 1960 N F f = µFN
F f = (0.25)(1960 N )
→
Fnet = Fpush − F f
→
→
→ FN = 1960 N F f = 490 N
Fnet = 1500.0 N − 490 N
→
Fnet = 1010 N
1010 N 200 kg
→
a = 5.05 m s2
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F = ma
→
a=
F m
→
a=
A 10,000 N elevator is supported by a cable that has a maximum tensile strength of 100,000 N. If the average passenger weighs 800 N, what is the maximum number of passengers the elevator can carry safely at an acceleration of 1.5 m/s2. given : weight elevator weight person
Since the elevator is accelerating (assume upward), the vertical forces are not balanced (Fnet ≠ 0). That means that the force of pull on the elevator has a larger magnitude than the force of gravity.
= 10,000 N = 800 N
Tmax
Fnet = Fpull – Fgravity
*note, force of gravity is negative in magnitude because it always acts opposite the pull upward
= 100,000 N a = 1.5 m s2
In this case we want to find Fpull because this will be the Tension in the cable.
needed equations : F
= ma → Fg = mg
You’ll need to find the mass of the elevator and the individual persons in order to calculate the Fnet.
Fg = mg → m =
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Fg g
Fg = mg → m =
Fg g
Fnet = Fpull − Fg
→
m(a + g) = T
→
m=
10000 N 9.8 m s2
→
m=
800 N 9.8 m s2
ma = T − mg
m=
→
m people =
T − melevator a+g
m people =
100,000N −1020kg 11.3 m s2
number of people =
→
T a+g
m people =
→
melevator = 1020 kg m person = 82 kg
→
→
→
→
ma + mg = T
(m people + melevator ) =
T a+g
100,000N −1020kg 1.5 m s2 + 9.8 m s2
m people = 8850kg −1020kg
masspeople avergage mass of one person
→
→
m people = 7830kg
number of people = 95.48
*A few things about this answer
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1. What?!? 95 is way too big an answer. The mass of the elevator should have probably been much larger than 1020 kg! 2. Also if you get an answer that is a fraction of a person, ROUND DOWN. 95.48 would become 95.
