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grr nother physics problem

M

MikeMike

Lifer
The coefficient of static friction between a block and a horizontal floor is 0.37, while the coefficient of kinetic friction is 0.1. The mass of the block is 6.7 kg. If a horizontal force is slowly increased until it is barely enough to make the block start moving, what is the net force on the block the instant that it starts to slide?

F = mn (m=mew thingy)

so we have a 6.7kg block or 6.7x9.8=65.66N

so now we have a .37 friction coefficient that you try to overcome,

or F= .37*65.66

but thats not the answer.

ive never been good with friction, just good at eliminating it 😀. any help?
 
if they mean net force in the horizontal direction it would be zero wouldn't it? the force you apply is equal to the static friction force.
 
Originally posted by: Yossarian
if they mean net force in the horizontal direction it would be zero wouldn't it? the force you apply is equal to the static friction force.
Click to expand...

its not 0 just tried that, i now have one more attempt to answer the problem, i have 5 tries, used 3 b4 i asked here.
 
.1*65.66N

hmm it should have been 65.66 * .37 not my .1 figure.

actually, after reading your problem, it should be something like .0000000001 since it asked for the NETforce. Unless they gave a force applied figure.
 
You need to calculate the amount of force needed to break the co-efficient of friction.

I think that's the phrase.

Wow, 10 year old data still stuck in my brain. Damn you beer! Do your work. :beer)
 
Originally posted by: chuckywang
Try subtracting off .1*N....I'm not really sure if you take the kinetic friction into account though.
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I would think at that instant it would be .37N - .1N = net force on the block in the horizontal direction.
 
Originally posted by: James3shin
kinetic friction doesn't matter, since the problem is just asking for the box to START moving
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That doesn't seem to be the answer they're looking for though, which is why we're suggesting others.
 
i hear what your saying Merlyn. I'm thinking this is one of those online HW assignments, and the Prof. may have made an error inputting acceptable answers. 🙂
 
Originally posted by: James3shin
i hear what your saying Merlyn. I'm thinking this is one of those online HW assignments, and the Prof. may have made an error inputting acceptable answers. 🙂
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its an online homework assignment, however iam pretty sure that the numbers are randomly generated and then solved by the program themselves given the proper formula, since they just change numbers.

would it possibly be: .37*9.81*6.7 + .1*9.81*6.7?
 
You're using a Masteringphysics type online homework system? Sometimes the questions are bugged. E-mail the professor.

And MAKE SURE that you're using the right number of sig figs in your final answer. I haven't used MP in over a year, but there were a few problems where the system would take an input and round it to 3-4 decimal places when the actual answer has no decimal places or maybe 2 decimal places of accuracy.

Try that, I think you should be good. The net force at the moment that the block starts to move is (m*g)*(mu_static) where mu_static=coefficient of static friction.

BTW, as an FYI, mu is often pronounced "mew" but the correct pronuniciation is "moo."
 
Originally posted by: Eeezee
You're using a Masteringphysics type online homework system? Sometimes the questions are bugged. E-mail the professor.

And MAKE SURE that you're using the right number of sig figs in your final answer. I haven't used MP in over a year, but there were a few problems where the system would take an input and round it to 3-4 decimal places when the actual answer has no decimal places or maybe 2 decimal places of accuracy.

Try that, I think you should be good. The net force at the moment that the block starts to move is (m*g)*(mu_static) where mu_static=coefficient of static friction.

BTW, as an FYI, mu is often pronounced "mew" but the correct pronuniciation is "moo."
Click to expand...

the programs name is CHIP.
 
Originally posted by: Eeezee
So in this case, 9.8*0.37*6.7 should be all you need. With sig figs, that's 24 with no decimals.
Click to expand...

it doesnt take sig figs into cause. it also gives about a 1.5% error difference like if i gave 24.15 instead of 24 i believe it would take it. there is some sort of error guidance it gives.
 
Yeah, that problem is really vague. Try F_friction_static - F_friction_kinetic as an answer. It should be 0 if they mean the net force right before moving, since it's not moving. At the instant that it starts moving, the acceleration is nearly infinitely small, so the net force should still be rounded off to 0. Sounds like a bad problem, e-mail the professor if it's not due tonight =p
 
yup that was it, to bad i tried + and not - so i got the answer right on my 6th attempt out of 5

o well only one problem
thanks guys. thats all for tonight, tune in again.
 
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