Finding the mass of a block on an inclined plane

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Homework Help Overview

The problem involves a block resting on an inclined plane, with a focus on determining the maximum mass that can be suspended without causing the block to slide down the incline. The scenario includes static friction and tension in a string over a pulley.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the forces acting on the block, including friction, tension, and weight. There are attempts to calculate the tension in the string and its relationship to the mass of the block hanging from it. Questions arise regarding the correct angles and the addition of forces.

Discussion Status

Some participants have made calculations regarding the forces involved and have shared their reasoning. There is an ongoing exploration of the correct setup and assumptions, with some guidance provided on the need to consider all forces acting on the block.

Contextual Notes

There are mentions of potential errors in angle usage and force calculations, indicating that participants are actively questioning their assumptions and the setup of the problem.

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Homework Statement


Block A with a mass of 10 kg rests on a 35degree incline. The coefficient of static friction is 0.40. An attached string is parallel to the incline and passes over a massless, frictionless pulley at the top. The largest mass mb, attached to the dangling end, for which A remains at rest is...?


Homework Equations


F = ma
Fs = us*Fn



The Attempt at a Solution


Fs = .4*10*9.8*cos35 = 32.1N

That's all I've got so far. :smile:

Not sure why I'm struggling so hard on this problem.
 
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hansel13 said:
Block A with a mass of 10 kg rests on a 35degree incline. The coefficient of static friction is 0.40. An attached string is parallel to the incline and passes over a massless, frictionless pulley at the top. The largest mass mb, attached to the dangling end, for which A remains at rest is...?
…
Fs = .4*10*9.8*cos35 = 32.1N

Hi hansel13! :smile:

Have you drawn an fbd for the block?

There's the friction (which you've caclulated), the tension, the normal force, and the weight.

So what is the tension? :smile:
 
lets calls the mass of block B, B for now.

I think I figured it out, so:
T = Fk-mg*sin30
= 32.1N -10kg * 9.8m/s2*sin30 = 24.1And we know that T = B*g

So B = T/g 24.1/9.8 = 2.46 kg

I'm pretty sure that's right, is it?

Thanks for the help! Lots of times I get lost and forget what I'm actually looking for, when you mentioned that I needed to find Tension I just drew it up and figured it out.
 
Hopefully it is haha..
 
If someone could let me know if this is right or not, I'd really appreciate it. I'm not too confident in my physics skills yet and this is an extra 5 points to my exam correct, so yeah...
 
hansel13 said:
T = Fk-mg*sin30
= 32.1N -10kg * 9.8m/s2*sin30 = 24.1

And we know that T = B*g

So B = T/g 24.1/9.8 = 2.46 kg

Hi hansel13! :smile:

Yes, that's more-or-less right, except …

i] isn't it 35º? :rolleyes:

ii] shouldn't you add the weight force to the friction force?
 
tiny-tim said:
Hi hansel13! :smile:

Yes, that's more-or-less right, except …

i] isn't it 35º? :rolleyes:

ii] shouldn't you add the weight force to the friction force?

ah, Right!

T = Fk-mg*sin30
= 32.1N + 10kg * 9.8m/s2*sin35 = 88.2


And we know that T = B*g

So B = T/g 88.2/9.8 = 9kg

Thanks! Dumb mistakes on my part..
 

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