# Homework Help: Friction, a sled, snow covered hill, sliding box

1. Oct 29, 2008

### Blangett

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

Question Details:
A rope attached to a 19.0 kg wood sled pulls the sled up a 18.0 kg snow-covered hill. A 9.0 kg wood box rides on top of the sled.

If the tension in the rope steadily increases, at what value of the tension does the box slip?

So far none of the answers I have tried has worked. I looked up the friction coefficients. μ=.50 (wood on wood)
μ=.12 (wood on snow)

I have a midterm I am studying for on Friday. I am trying to grasp these types of problems unsuccessfully. Any help you can provide is deeply appreciated.

2. Relevant equations

Where do I go from here? I am stuck on how to proceed. The fact that I am using the net forces of two seperate objects is confusing me. I am not sure what to do with them (do I add the Fnet x together?. Also the fact that I do not have acceleration given is bothering me.

3. The attempt at a solution

Known
fk=μkn
ms=19.0kg
mb=9.0kg
μk=.06 (wood on snow)
μs=.50 (wood on wood)
n=m*g
gx=9.8sinθ m/s2
gy=9.8cosθ m/s2

(Fnetx)sled=T-fk-g
= T-μkn-msg*sinθ=msax
= T-μkmsg *cosθ-msg*sinθ=msax
(Fnety)sled = n-g=msax
= n-msg*cosθ=m*0
n =ms*g*cosθ

(Fnetx)box = -fs-g = mbax
=μs*mb*g*cosθ*mb*g*sinθ =mbax
(Fnety)box= n - g
= n - mb*g*cosθ=mbay
= n - mb*g*cosθ=0
n= mb*g*cosθ

2. Oct 30, 2008

### LowlyPion

Welcome to PF.

I presume that θ = 18 and that it is not an 18 kg hill.
Also I'm not sure whether your coefficients of friction are given as part of the problem or you are supplying them. But you haven't properly recorded it in your equations.

Without working through your equations, let me just say generally that the first thing to find is the acceleration of the system (box and sled) - once T overcomes wood/snow friction of both normal components of weight as well as the component of weight down the incline. That acceleration then applies to both masses - (sled + box).

Armed with acceleration then at what point does the acceleration against the box overcome the normal component of the weight times the wood wood coefficient?

3. Oct 24, 2010

I'm working on this same problem, and rather than post a new thread, I thought it best to continue on this one. So far I've gotten just half a step father than the OP. First I'll restate the problem in general terms:

A rope attached to an $$M_{s}$$ kg wood sled pulls the sled up a θ-degree snow-covered hill. An $$M_{b}$$ kg wood box rides on top of the sled. If the tension in the rope steadily increases, at what value of the tension does the box slip? I'm given $$M_{s}$$, $$M_{b}$$, θ, and both static and kinetic μ of friction for wood-on-wood, and wood-on-snow.

First thing I did was make a free-body-diagram of the box, the sled, and of them both as a system. Then I double checked the quantity of forces by drawing an interaction diagram.
The box has 3 forces: weight, normal, and friction.
The sled has 6 forces: tension, weight, friction from ground, normal from ground, friction from box, & normal from box.
As a system there are 4 forces: normal, tension, weight, and friction.

I set up the hill as my (+) x-axis, and the y-axis perpendicular to that. Thus the weights are decomposed into:
$$\vec{w}$$=(-)(M)(sin(θ))$$\hat{x}$$ + (-)(M)(cos(θ))$$\hat{y}$$

Using the system, when the sled is just about to start moving: T - $$F_{ss}$$ - $$M_{t}$$ = 0
so T = ($$M_{t}$$)(g)(sin(θ)) + (μ_ss)($$M_{t}$$)(g)(cos(θ))
Then I divide by M to get A, because [T = MA] ==> [A = T/M]. This gives:
A = (g)( sin(θ) + (μ_ss)cos(θ) )

Now, I'm unsure what is meant by "acceleration against the box." I guess my problem is that I don't understand how the tension on the sled relates to the friction on the box.

In my FBD of the sled, both frictional forces are pointing in the (-) x-direction. So I figured that when the weight of the box is equal to the frictional force on the box it will be just about to slip. So I put the maximum static friction of the box into the equation of the sum-of-forces for the sled:
T - ( $$f_{sled}$$ + $$f_{box}$$ ) - ( ($$M_{s}$$)(g)(cos(θ)) ) = 0
T - [ (μ_ss)($$M_{s}$$)(g)(cos(θ)) ] - [($$M_{b}$$)(g)(sin(θ))] - [($$M_{s}$$)(g)(sin(θ))]
Solving for T gives does not give me the correct answer. I know, from the back of the book, that when $$M_{s}$$ = 20kg, $$M_{b}$$ = 10kg, and θ= 20-degrees, that T = 155N.

I tried something else too, but it's all gibberish now.