Friction and toboggans

mardi 30 juillet 2013

1. The problem statement, all variables and given/known data

A 15kg toboggan is pulled across the snow at a constant speed by a horizontal force of 22N.

a) What is the force of gravity on the toboggan?

b) What is ##μ_k##?

c) How much more force is needed to pull the toboggan if 105kg is added to the toboggan?

2. Relevant equations

##m = 15kg##

##a = 9.8 m/s^2##

##F_A = 22N## [Applied force]

##F_N## [Force normal]

##F_K## [Kinetic force/Friction]

3. The attempt at a solution

a) ##F_G = ma = (15 kg)(9.8 m/s^2) = 147N##

b) The toboggan travels at a constant velocity ( a = 0 ) with no ups and downs. This implies that ##F_N = F_G = 147N## and ##F_K = F_A = 22N##

Now ##F_K = μ_KF_N \Rightarrow μ_K = \frac{F_K}{F_N}##

So that ##μ_K = \frac{22N}{147N} = 0.15##.

c) Hm so the total weight of the toboggan will now be ##m = 120kg## and we have the kinetic friction coefficient ##μ_K = 0.15##.

Would it just be :

##F_K = μ_Kma = (0.15)(120 kg)(9.8 m/s^2) = 176.4N##

So we would require 176.4N+22N = 198.4N more to keep the toboggan at a constant velocity. Is this okay?