Finding the mass of a block on an inclined plane

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A block with a mass of 10 kg is on a 35-degree incline, with a static friction coefficient of 0.40. The user calculated the static friction force as 32.1 N but struggled with determining the tension in the string. After guidance, they corrected their calculations to include the weight force and found the tension to be 88.2 N. This led to the conclusion that the largest mass attached to the pulley, for which the block remains at rest, is 9 kg. The discussion highlights the importance of careful consideration of forces in physics problems.
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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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