# Homework Help: Person dragging a bow with a rope - Calculate friction

1. Dec 6, 2009

### dois

1. The problem statement, all variables and given/known data
The question says
"A person is dragging a box with the mass of 60kg by a rope. The person begins to drag the box at 4 m/s [F]. All of a sudden the person sees on coming traffic. The person calculates that they have 3 seconds to get out of the way. If at the end of the 3 seconds the box is moving 10 m/s [F] and the coefficient of friction between the box and the ground is 0.7, find the force which the person is pulling the box."

my biggest problem is I don't know what a coefficient is.

2. Relevant equations
im not to sure what equations i would need to use. My guess would be d=(vi)(t) + 1/2 (a)(t^2)

3. The attempt at a solution
To find out the force which the person is pulling the box would we still use
F= m*a

and we would need to figure out the displacement first with d=(vi)(t) + 1/2 (a)(t^2)

2. Dec 6, 2009

### Fightfish

Re: Forces

The coefficient of friction, $$\mu$$ is the constant relating the frictional force between two surfaces to the normal contact force between the two surfaces.
$$F_{friction} = \mu N$$​

The velocity of the box increases from 4m/s to 10m/s in 3s. I presume that should be enough for you to obtain the average force exerted on the box during this time period using the standard F=ma?

3. Dec 6, 2009

Re: Forces

so μ is 0.7?

4. Dec 6, 2009

Re: Forces

yep.

5. Dec 6, 2009

### dois

Re: Forces

How would we begin this question
my known's are
mass= 60kg
vi=4 m/s [F]
v2=10m/s [F]
t= 10s
friction=0.7

but i dont know what equation to use to find the applied force

6. Dec 6, 2009

### wisvuze

Re: Forces

you have time, v0 and vf. from there, you must solve for acceleration, then plug it into F = ma , which will be your net force. From there, you must consider all the different forces and single out the pulling force. (Hint: the pulling force is not all that is affecting the pull.. what does friction do?)

7. Dec 6, 2009

### wisvuze

Re: Forces

friction is not 0.7, that's the coefficient of friction. Friction is the coefficient of friction multiplied by the normal force

8. Dec 6, 2009

### dois

Re: Forces

ok, and the normal force is equal to the mass right?
so then it would be
0.7 x 60
which would then be
42?

9. Dec 6, 2009

### wisvuze

Re: Forces

the normal force is what's keeping the object "up", it is the sum of vertical components.. so in this example, you only have the force of gravity or the weight pushing it down to the ground, so the normal force is equal to the weight

10. Dec 6, 2009

### dois

Re: Forces

Alright. So then the normal force is 60, because the weight is 60kg.
and then friction would be, 0.7x60=42
is that right?

11. Dec 6, 2009

### wisvuze

Re: Forces

no, mass and weight are two different things. I think that you should revise your concepts