# Friction and toboggans

1. Jul 30, 2013

### Zondrina

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?

2. Jul 30, 2013

### PeterO

Are you sure you should add 22N at the end.

Have you calculated the force needed for the loaded toboggan, or the the extra force needed for the load?

3. Jul 30, 2013

### barryj

I think PeterO says it all

4. Jul 30, 2013

### Zondrina

You're right. My first instinct was 176.4N was the answer. Then I second guessed myself for a moment.

So we would need 176.4N to pull to toboggan.

5. Jul 30, 2013

### barryj

Be careful of the question. It asks for how much MORE force is needed, not how much force is needed.

6. Jul 30, 2013

### Zondrina

Then I would need 176.4N - 22N = 154.4N MORE force to pull the toboggan.

176.4N Is the force needed to pull the toboggan while 154.4N is how much more force I need to move the toboggan as compared to when the toboggan didn't have as much mass ( It only required 22N ).

7. Jul 30, 2013

### barryj

Correct

8. Jul 31, 2013

### haruspex

You seem to have introduced an error of 0.4N. 105kg is exactly 7 times 15kg, so the extra force should be exactly 7 times 22N.