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

  • #1
onemic
25
2

Homework Statement


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.

lesson 3.png
Given: m1=20kg, m2=10kg, μ=0.4

Required: Ffm1, Ffm2

Homework Equations


Fg=mg
FNet=ma
Ff=μFN

The Attempt at a Solution



lesson 3_2.png
Let the directions below equal positive:

lesson 3_3.png


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 I am not sure what to do as it doesn't seem like there's any way to solve for theta due to multiple variables...Is there an important step I'm missing?
 
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  • #2
onemic said:

Homework Statement


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.

View attachment 106658Given: m1=20kg, m2=10kg, μ=0.4

Required: Ffm1, Ffm2

Homework Equations


Fg=mg
FNet=ma
Ff=μFN

The Attempt at a Solution



View attachment 106660Let the directions below equal positive:

View attachment 106661

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 I am not sure what to do as it doesn't seem like there's any way to solve for theta due to multiple variables...Is there an important step I'm missing?
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##?
 
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  • #3
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
onemic said:
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

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!
 
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  • #5
PeroK said:
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!

hm, is it:

1=196sinθ/78.4cosθ
1=2.5tanθ
θ=tan-1(2.5)
θ=68.2
 
  • #6
That's the right idea, but your basic algebra has let you down.
 
  • #7
PeroK said:
That's the right idea, but your basic algebra has let you down.

Wow, I don't 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
onemic said:
Wow, I don't 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?

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?
 
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  • #9
The fact that friction is modeled, 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. [emoji4]
 
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  • #10
It's equal to the static force!
PeroK said:
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?
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!
sophiecentaur said:
The fact that friction is modeled, 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. [emoji4]

Thanks for the further explanation
 

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