# How much mass is needed to prevent an object from sliding down a ramp

1. Jul 8, 2014

### physicsgirl199

1. The problem statement, all variables and given/known data
An 80 kg object is on top of a 40°ramp and held by a box that is filled with sand bags. Each sand bag has a mass of 12 kg, the static friction coefficient between the box and the ramp is .507. What is the least amount of sand bags needed to prevent the object from sliding.

2. Relevant equations
F,net,,x,object = 0 = F(box) - F(grav,x)
F(box) = F(grav,x,obj)

F,net,box = friction - F(object) - F(grav,x,box)
friction = μ(s)mgcos(40°)
F(grav,x,box) = mgsin(40°)
F(object) = mg(sin°40)

3. The attempt at a solution

F(box) = F,grav,x,obj = 504 N

F,net,box = friction - F(obj) - F,grav,x,box
= > μ(s)mg(cos40) - 504 - mg(sin40) = 504
=> m[μ(s)gcos40 - gsin40] = 1008
=> m = 1008/(-2.49) = -404 kg ??

so lost

2. Jul 8, 2014

### BiGyElLoWhAt

Is there no friction between the 80kg object and the ramp?

3. Jul 8, 2014

### Bandersnatch

This looks like a trick question(providing all the data is correct).

Perhaps if you calculate the net force acting on the box parallel to the ramp, completely disregarding the 80kg body, you'll see why.

4. Jul 8, 2014

### physicsgirl199

There is no information regarding static friction between the object and ramp..

If i disregard the object and focus on the box that is on the ramp, wouldnt the net force be 0 with friction and F,g,x cancelling each other out if it does not move?

5. Jul 8, 2014

### BiGyElLoWhAt

As bandersnatch pointed out, just do a N2L for the box with the sand bags and mass M, ignore the 80kg object, and see what happens.

6. Jul 8, 2014

### physicsgirl199

F box = friction - F,gx --> u(s)mgcos40 - mgsin40 = 0 --> factor out m --> m = 0??

7. Jul 8, 2014

### BiGyElLoWhAt

If you factor out the m (and g) from your final eqation you get
$mg(\mu_{s}cos(40) - sin(40)) = ma = 0$
If you plug in your known values, what does that tell you? (you're on the right track, just think about the implications)

Try dividing.

8. Jul 8, 2014

### physicsgirl199

divide out m and your left with a negative acceleration meaning that box is moving meaning friction would not keep the box up on the ramp no matter what?

9. Jul 8, 2014

### Bandersnatch

Looks that way.

10. Jul 8, 2014

### BiGyElLoWhAt

Basically yea.

$\mu_{s}\text{cos}(40)$ is always less than $\text{sin}(40)$ for the given value of mu s.

11. Jul 9, 2014

### verty

I calculate that there is an answer only when Θ < 26.885°.

12. Jul 9, 2014

### BiGyElLoWhAt

That could be, I didn't go that far, but you need mu s cos theta >= sin theta which doesn't happen at 40 degrees.