Find the angle of an object on an incline (static friction)

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1. Sep 29, 2016

onemic

1. The problem statement, all variables and given/known data
At a construction site, a small crane is raising two boxes of nails on a plank to the roof. One box has already been opened and is half full, while the other box is new. The boxes, including the nails weigh 10kg and 20kg, respectively, and are the same size.
If the coefficient of static friction is 0.4, draw an FBD for each box of nails and use it to calculate the angle at which each box begins to slide.

Given: m1=20kg, m2=10kg, μ=0.4

Required: Ffm1, Ffm2

2. Relevant equations
Fg=mg
FNet=ma
Ff=μFN

3. The attempt at a solution

Let the directions below equal positive:

M1:

FNety=Fgy-FN
0 = Fgy-FN

Fgy=FN
(mg)cosθ=FN
(20kg)(9.8N/kg)cosθ=FN
FN=196cosθ

FNetx=196sinθ-μFN
=196sinθ-(0.4)196cosθ
=196sinθ-78.4cosθ

This is where I'm at currently and Im not sure what to do as it doesnt seem like there's any way to solve for theta due to multiple variables....Is there an important step I'm missing?

2. Sep 29, 2016

PeroK

As far as I can see you are nearly there. What can you say about $F_{net}$ when the box is about to slip?

By the way, do you think the angle of slippage depends on $g$ or $m$?

3. Sep 29, 2016

onemic

FNet is equal to 0 when the box is about to slip due to there being no acceleration.

0=196sinθ-78.4cosθ
78.4cosθ=196sinθ

The angle of slippage would depend on m, since m is variable while g is constant. The higher the value of m, the greater the angle the platform would have to be for the box to slip

4. Sep 29, 2016

PeroK

You should be able to solve your equation for $\theta$. Is there another common trig function, other than $\sin$ and $\cos$?

On the second point, I'll be interested in the difference in $\theta$ for the two boxes, once you've calculated them!

5. Sep 29, 2016

onemic

hm, is it:

1=196sinθ/78.4cosθ
1=2.5tanθ
θ=tan-1(2.5)
θ=68.2

6. Sep 29, 2016

PeroK

That's the right idea, but your basic algebra has let you down.

7. Sep 29, 2016

onemic

Wow, I dont know how I missed that.

1=2.5tanθ
1/2.5=tanθ
θ=tan-1(1/2.5)
θ=21.8

The second box...also has an angle of 21.8 degrees. Very surprising. So the angle of slippage is dependent on g and not on m? Or is it entirely dependent on the static force?

8. Sep 30, 2016

PeroK

You could try calculating the angle for a different value of $g$. The moon's gravity is about $1.6 m/s^2$ I think.

Also, do you notice anything about the number $1/2.5$? Is that equal to anything in the problem?

9. Sep 30, 2016

sophiecentaur

The fact that friction is modelled, using a 'coefficient' suggests that the ratio of parallel and normal forces is considered to be independent of the actual values.
This is an 'ideal' model, of course and there are limits to the range of values of the forces. But your calculation makes this assumption from the start. You have actually just shown that 1/x =1/x.

10. Sep 30, 2016

onemic

It's equal to the static force!
It's equal to the coefficient of static friction! So it's the static friction coefficient that determines the angle that the boxes will slip. That was a trick question :p

Thanks for the help!
Thanks for the further explanation