Determining the angle for a mass to overcome friction

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Nojins
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Homework Statement


A crane is moving two equally spaced masses to the roof. One box is 10kg and the other box is 20kg. As the plank tilts towards the 20kg box, at what angle will each box begin to slide? The coefficient of static friction is 0.4

Homework Equations


Newton's second law, Fnet=ma

The Attempt at a Solution


Originally I thought I had correctly solved it, by simply setting Fnet=0 and solving for theta, however I didn't realize that the force of friction also depended on the angle, which I hadn't solved for. I ended up with this and I'm unsure of how to continue. Am I also correct in assuming both boxes will move at the same time?
Fnet=mgsinθ-Ff
0=10(9.8)sinθ-0.4(mgcosθ)
Thanks.
 
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Nojins said:

Homework Statement


A crane is moving two equally spaced masses to the roof. One box is 10kg and the other box is 20kg. As the plank tilts towards the 20kg box, at what angle will each box begin to slide? The coefficient of static friction is 0.4

Homework Equations


Newton's second law, Fnet=ma

The Attempt at a Solution


Originally I thought I had correctly solved it, by simply setting Fnet=0 and solving for theta, however I didn't realize that the force of friction also depended on the angle, which I hadn't solved for. I ended up with this and I'm unsure of how to continue. Am I also correct in assuming both boxes will move at the same time?
Fnet=mgsinθ-Ff
0=10(9.8)sinθ-0.4(mgcosθ)
Thanks.
Let's start out with a clearer description of the problem. Is the idea that two boxes are placed on a plank and the plank is tilted until the boxes start to slide? I wonder whether the crane or the roof have anything to do with the problem?
 
Nojins said:
Am I also correct in assuming both boxes will move at the same time?
I'm unsure how hard the problem is supposed to be.
The rotation of the plank will be accelerating. This will affect the normal forces.
Also, the rate of rotation will affect the radial forces. The lower block should slip first.
 
haruspex said:
I'm unsure how hard the problem is supposed to be.
The rotation of the plank will be accelerating. This will affect the normal forces.
Also, the rate of rotation will affect the radial forces. The lower block should slip first.
Probably they are just describing a crane with a plank capable of tilting. And motion/rotation is not considered (not enough info is given to solve for these).
So basically it is just: you have a plank and you start to tilt it. What angle would a 10 kg mass slide, and then the same situation with a 20kg mass.
 
Nojins said:
Am I also correct in assuming both boxes will move at the same time?
Fnet=mgsinθ-Ff
0=10(9.8)sinθ-0.4(mgcosθ)
Thanks.
Do not assume that they slide at the same time. They tell you that the masses are spaced. Calculate each angle separately (an angle for 10 and an angle for 20).
Can you rearrange this: 0=10(9.8)sinθ-0.4(mgcosθ) into something easier to find theta? Perhaps a single trigonometric function, rather than two functions?
 
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scottdave said:
motion/rotation is not considered (not enough info is given to solve for these).
I agree that a simple treatment is probably intended, but there is likely enough information for the thorough approach. Just consider the plank as initially level. The only datum we do not have is the horizontal separation (x).
If we say that each slip angle (if the other has not slipped first) is a function of mass ratio, g, x, and friction coefficient then dimensional analysis rules out g and x.
 
haruspex said:
I agree that a simple treatment is probably intended, but there is likely enough information for the thorough approach. Just consider the plank as initially level. The only datum we do not have is the horizontal separation (x).
If we say that each slip angle (if the other has not slipped first) is a function of mass ratio, g, x, and friction coefficient then dimensional analysis rules out g and x.
Having gone through the approach of finding the angular velocity and acceleration as functions of θ, it turns out not that hard and the result is quite elegant. So now I believe this is the intended solution.
 
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