What Is the Coefficient of Friction for a Toboggan with Added Weight?

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The force of gravity acting on a 12kg toboggan is calculated to be 118N. The coefficient of friction for the toboggan is determined to be 0.093 when pulled with an 11N force at constant velocity. When two 57kg girls sit in the toboggan, the total mass increases to 126kg, resulting in a normal force of 1234.8N. To maintain constant velocity with the added weight, the same coefficient of friction can be used to find the required horizontal force. The discussion emphasizes the relationship between force, mass, and friction in maintaining constant motion.
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A 12kg toboggan is pulled along at a constant velocity on a horizontal surface by a horizontal force of 11N.
a) What is the force of gravity on the toboggan?
b) What is the coefficient of friction?
c) How much horizontal force is needed to pull the toboggan at a constant velocity if two 57kg girls are sitting in it?

m=12kg
FA= 11N

a) Fg=FN
mg=FN
12(9.8)=FN
118N=FN=Fg

Fg=118N

b) FA=Ff; constant vel.
FK=FA=11N

Fk=μKFN
11N= μK(118N)
μK=11N/118N
μK=0.093

c)m= 12kg + 2(57kg)= 126kg

Fg=FN
mg=FN
126(9.8)=FN
1234.8N=FN


m=126kg
FN=1234.8N
FK=FA=?
im stuck here..
 
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xdeanna said:
A 12kg toboggan is pulled along at a constant velocity on a horizontal surface by a horizontal force of 11N.
a) What is the force of gravity on the toboggan?
b) What is the coefficient of friction?
c) How much horizontal force is needed to pull the toboggan at a constant velocity if two 57kg girls are sitting in it?

m=12kg
FA= 11N

a) Fg=FN
mg=FN
12(9.8)=FN
118N=FN=Fg

Fg=118N

b) FA=Ff; constant vel.
FK=FA=11N

Fk=μKFN
11N= μK(118N)
μK=11N/118N
μK=0.093

c)m= 12kg + 2(57kg)= 126kg

Fg=FN
mg=FN
126(9.8)=FN
1234.8N=FN


m=126kg
FN=1234.8N
FK=FA=?
im stuck here..

You have the coefficient of friction from part b. You can use this value to find the force of kinetic friction acting on the toboggan+kids.

And because the tobaggon must move at a constant velocity, the net force must be zero. Thus, like you wrote, Fa = Fk.
 
Ahalp said:
You have the coefficient of friction from part b. You can use this value to find the force of kinetic friction acting on the toboggan+kids.

And because the tobaggon must move at a constant velocity, the net force must be zero. Thus, like you wrote, Fa = Fk.

omg i feel stupid now :|
thanks lol
 
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