Another Physics Problem

fusilli_jerry89 · 2006-10-04 10:59:23

Sorry for posting another physics problem here, but you are the only ones that actually help.

The drawing shows a large cube(mass=25 kg) being accelerated accross a frictionless surface by a horizontal force P. A small cube(mass=4.0 kg) is in contact with the front surface of the large cube and will slide down unless P is sufficiently large. The coefficient of static friction between the cubes in 0.71. WHat is the smallest magnitude that P can have in order to keep the small cube from sliding downward?

I understand that an extra 11.368 Newtons must be directed upwards to keep the block in place but I don't know what to do after that.

John E. Franklin · 2006-10-04 12:00:53

Here's some reading:
http://courses.ncssm.edu/physics/ILD/co … ect_fo.htm

John E. Franklin · 2006-10-04 12:24:48

Some more reading:
http://www.dac.neu.edu/physics/m.glaubman/a03.pdf

John E. Franklin · 2006-10-04 13:07:30

Reading on something that is often neglected in high school, but talked about in college.
Free body diagrams; a diagram with forces shown for each object in the diagram.
Normal forces where objects touch each other. Gravity, applied forced like pulling object...
http://a-s.clayton.edu/campbell/physics … 1111fb.htm

John E. Franklin · 2006-10-04 13:17:45

Figure out the acceleration of the combined 29 kgs (25 + 4).
a = P/29
Then F=ma for normal force ( I think?) touching between the two cubes.
A force acts rightward to move the tiny cube at a = P/29 acceleration.
A force acts leftward back on the big object equal to that rightward force I think perhaps.
This force on the tiny cube is F=4(P/29), I think, maybe, it's been a long time since I did this.
Then use friction ratio of .71 multiplied by either F or a, however that works, I can't remember.
This gets you the opposing static friction force upward against gravity.

John E. Franklin · 2006-10-04 14:15:11

So probably, jumping ahead a bit, 9.8067 / .71 is the rightward acceleration needed or faster.
So the 4 kg cancels out, and it doesn't matter how heavy it is I think.
And if P/29 > 9.8067 / .71, then it stays up and doesn't slide, maybe, just maybe!!

fusilli_jerry89 · 2006-10-17 17:03:31

wat did u do between your last two posts? bcuz u got the right answer

Last edited by fusilli_jerry89 (2006-10-17 17:03:47)

fgarb · 2006-10-18 00:50:34

Hi Fusilli,

If I may add my own suggestion here, the first thing I recommend when doing these problems is to draw a free body diagram. Draw the two boxes, and write arrows with every force you can think of labeled on them. This will both aid your thinking and encourage the grader to give you more partial credit if you make a mistake solving the problem. In this case, your free body diagram should have the force P, the normal forces of the boxes pushing on each other, the force of friction, and the force of gravity.

It's also much easier to keep track of things if you wait to plug in actual numbers until the very end. That makes it way easier to find mistakes and will also help you get more partial credit. Let me call the masses M (for the big one), m (for the small one), and F_sub_N the normal force. The condition we want to solve is:

[align=center]

[/align]
[align=center][/align]

You know that the normal force is the force that's accelerating the little box to the right. Since force is just mass times acceleration,

[align=center]

[/align]

So you can plug that in, but you still don't know what the acceleration is. So you have to use one last piece of information. You know that the overal pushing force (I'll call it F-sub-p) has to be just right to accelerate both masses together at a. So,

[align=center]

[/align]

You can plug the normal force back in to get

[align=center]

[/align]
[align=center][/align]

If you plug in all the values, you will see this is the same result as John got, and I think this is basically the same reasoning as he used. Please ask if anything about this is confusing!

fusilli_jerry89 · 2006-10-24 18:08:26

what is the difference between the normal force and what you call Fp?

George,Y · 2006-10-24 20:14:38

the difference lies in which object they act on.

fusilli_jerry89 · 2006-10-25 12:05:42

so basically they are the same but one is negative?

fgarb · 2006-10-25 17:00:07

I was using F-sub-p in place of P in your diagram to represent the horizontal force being applied to the large block. I probably should have stuck with P to be less confusing, but P is often used to represent pressure, so I instinctively avoid that. It is not the same as the normal force F-sub-N being applied between the two blocks. I will be a little more explicit here:

The small block is accelerated by the normal force F-sub-N, so

[align=center]

[/align]

Where the small block has mass m and is being accelerated horizontally with acceleration a. But we know that the big block is also accelerated at exactly a too (otherwise they wouldn't always be touching). So F-sub-p (or "P" the way you were using it) is

[align=center]

[/align]

Since it has to push both masses at acceleration a. If you need a better justification for this let me know and I can give it to you but I'm tired now, so I'll go on to the punchline. I then solved for a and plugged it back into the F-sub-N equation:

[align=center]

[/align]

The stuff inside the {} is just a. Does this all make sense?

George,Y · 2006-10-25 17:32:00

It's like somebody behind you just push you, and you push someone in front of you, and they are not the same push.

fusilli_jerry89 · 2006-12-18 18:21:34

could u simply solving this problem by seeing that 39.2 newtons must be supplied upwards(Fg) and dividing that by mu to get 55 N as the normal force. You can then see than the big box is 6.25 times more mass and therefore requires 6.25 times more force. 6.25 times 55 +55. You get the same answer only it's a hell of a lot easier. I get the other way too tho.

Stanley_Marsh · 2006-12-19 02:01:07

the horizontal accelerations of two cubes are the same~ if regard the two as one , we have a=p/29he large cube endures 25*(p/29)wton force , this force equals to the sum of the p force and the force the small one attributes . the interactiing force measured p-(25p/29) , the friction = [p-(25p/29)]*0.71 when the friction equals to the gravity of the smalls one , it will be minimum , [p-(25p/29)]*0.71=4*9.8 , which p is around 400 N ,I dont know if I am correct lol~

fgarb · 2006-12-19 05:53:38

fusilli, what you say is true, and if you get the same answer that way, that might be a good way to think about it. I think you're right too Stanley -- at least you're expression for the normal force looks like it matches mine.

I will say that the reason these approaches look simpler though is because you are combining a number of steps into one. As long as you are careful that's fine, but it is much easier to make mistakes that way and it's harder to justify your steps to others. The most organized and convincing way (in a physics teacher's point of view) to answer a problem like this:

1) Draw a free body diagram and label all the relevant forces
2) Write down the equations for the forces and accelerations you want to use
3) Use these equations step by step to show how you get your answer
4) Plug in numbers for your symbols at the end (sure you can do it earlier, but it's much easier to grade and you don't have to use your calculator until the end this way, so there are fewer oportunities for mistakes)

Math Is Fun Forum

#1 2006-10-04 10:59:23

Another Physics Problem

#2 2006-10-04 12:00:53

Re: Another Physics Problem

#3 2006-10-04 12:24:48

Re: Another Physics Problem

#4 2006-10-04 13:07:30

Re: Another Physics Problem

#5 2006-10-04 13:17:45

Re: Another Physics Problem

#6 2006-10-04 14:15:11

Re: Another Physics Problem

#7 2006-10-17 17:03:31

Re: Another Physics Problem

#8 2006-10-18 00:50:34

Re: Another Physics Problem

#9 2006-10-24 18:08:26

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#10 2006-10-24 20:14:38

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