very basic physics question (thats driving me nuts)
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very basic physics question (thats driving me nuts)
there are two parts to the question: find the acceleration of the system, and find the force the two objects exert on each other.
I found the acceleration fine and dandy, no problem, it's about 2.5 m/s^2. But the second part is baffling me for some reason. The answer stated by my textbook is 440 N, but all of the things Im trying are not coming up with that number. First I just multiplied the mass of the object on the left by it's acceleration (found in the first part of the question), but it does not produce 440 N. Then I took 730 N force vector and subtracted the force of friction (working on the left object) to find the resultant force.... still no beans. I'm terribly confused because it seems like such an elementary problem but I can't figure it out Can someone please explain to me why force exerted by the blocks on each other is 440 N?
Re: very basic physics question (thats driving me nuts)
u as in the coefficient of static friction?
What's acting on the box on the left? From your diagram, it dosen't look like there would be any forces acting on it (other than normal and gravitational, which would cancel)
Are the boxes attached by a rope or something to that effect?
I presume that 730N is from an external pulling force?
What's acting on the box on the left? From your diagram, it dosen't look like there would be any forces acting on it (other than normal and gravitational, which would cancel)
Are the boxes attached by a rope or something to that effect?
I presume that 730N is from an external pulling force?
 skeptical scientist
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Re: very basic physics question (thats driving me nuts)
Can you please say what the question said, word for word from the textbook? Right now it seems like only a halfformed question.
I'm looking forward to the day when the SNES emulator on my computer works by emulating the elementary particles in an actual, physical box with Nintendo stamped on the side.
"With math, all things are possible." —Rebecca Watson
"With math, all things are possible." —Rebecca Watson
Re: very basic physics question (thats driving me nuts)
I assume the 730N force is pushing horizontally to the right on the 75kg block.
Since you already have the acceleration, you can find the net force on the 110kg block. You can also find the friction force on the 110kg block. Then you must determine what the applied force must be to overcome friction and result in the net force. That is the force applied by the 75kg block on the 110kg block.
Do you know how to determine the force applied by the 110kg block on the 75kg block?
Since you already have the acceleration, you can find the net force on the 110kg block. You can also find the friction force on the 110kg block. Then you must determine what the applied force must be to overcome friction and result in the net force. That is the force applied by the 75kg block on the 110kg block.
Do you know how to determine the force applied by the 110kg block on the 75kg block?
wee free kings
Re: very basic physics question (thats driving me nuts)
You use ratios for this question.
Basically you add up the total mass of the system, in this case 185 (75 + 110).
Then you find out what percentage of the mass is composed of either block by dividing the mass of a single block by the mass of the system:
75 kg block: 75 / 185 = 0.405
110 kg block: 110 / 185 = 0.595
Now, multiplying those decimals by the total force on the system:
75 kg block: 0.405 * 730 N = 300 N
110 kg block: 0.595 * 730 N = 430 N
I've adjusted the results according to significant figures. The discrepancy is probably a result of rounding errors somewhere.
Basically you add up the total mass of the system, in this case 185 (75 + 110).
Then you find out what percentage of the mass is composed of either block by dividing the mass of a single block by the mass of the system:
75 kg block: 75 / 185 = 0.405
110 kg block: 110 / 185 = 0.595
Now, multiplying those decimals by the total force on the system:
75 kg block: 0.405 * 730 N = 300 N
110 kg block: 0.595 * 730 N = 430 N
I've adjusted the results according to significant figures. The discrepancy is probably a result of rounding errors somewhere.
Re: very basic physics question (thats driving me nuts)
sorry for being so vague in my question in the diagram, u is coefficient of kinetic friction, and the 730 N force is pushing the entire system to the right.
this is the straight text of the question: Two crates, of mass 75 kg and 110 kg, are in contact and at rest on a horizontal surface. A 730N force is exerted on the 75kg crate. If the coefficient of kinetic friction is .15, calculate (a) the acceleration of the system, and (b) the force that each crate exerts on the other.
this is the straight text of the question: Two crates, of mass 75 kg and 110 kg, are in contact and at rest on a horizontal surface. A 730N force is exerted on the 75kg crate. If the coefficient of kinetic friction is .15, calculate (a) the acceleration of the system, and (b) the force that each crate exerts on the other.

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Re: very basic physics question (thats driving me nuts)
Yambert wrote:
there are two parts to the question: find the acceleration of the system, and find the force the two objects exert on each other.
I found the acceleration fine and dandy, no problem, it's about 2.5 m/s^2. But the second part is baffling me for some reason. The answer stated by my textbook is 440 N, but all of the things Im trying are not coming up with that number. First I just multiplied the mass of the object on the left by it's acceleration (found in the first part of the question), but it does not produce 440 N. Then I took 730 N force vector and subtracted the force of friction (working on the left object) to find the resultant force.... still no beans. I'm terribly confused because it seems like such an elementary problem but I can't figure it out Can someone please explain to me why force exerted by the blocks on each other is 440 N?
You should not try every formula possible to make the result come up. The point of solving a problem is to understand it, not randomly putting numbers. If you solve problems this way, not only you are gonna be frustrated in the process, but also you are getting very little or no results. You should not care about the solution your books gives you until you fully solved the problem.
In your case you should draw a freebody diagram for each of the two boxes, displaying all the forces etc.
Then you examine all the forces acting on the first block, then all the forces acting on the second block.
Since they have the same acceleration, indicate it with a, and since the contact force between them is the same in magnitude, you can call it Fc.
You know that
Fnet_1 = m_1a
Fnet_2 = m_2a
From these two equations you can determine both Fc and a.
 Sableagle
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Re: very basic physics question (thats driving me nuts)
Starting with the assumptions that:
the two crates are in facetoface contact, not edgetoface or edgetoedge contact;
the two crates are cuboids;
the two crates are each resting on a face, with neither resting on an edge and the other crate;
the coefficient of friction provided is between crate and surface;
friction is unaffected by velocity (because they didn't specify only initial acceleration);
the horizontal force specified is pushing the 75 kg crate against the 110 kg crate in a direction normal to their faces in contact;
the horizontal force specified is pushing in line with the centres of mass of both crates;
the horizontal force specified is pushing in line with the centres of friction under both crates;
air (or other fluid) resistance is negligible at the speeds involved (well, it does say "at rest" but if they're under treacle this may be an issue);
the two crates are of similar enough size for neither to topple over the other
and
the crates contents are rigid and cannot move around within the crates ...
Thing to check number 1: total friction under the crates is 0.15 * ( 75 + 110 ) * g, which is 271.0065 N in Singapore, 272.64375 N in Helsinki or 45.0105 N on the Moon. Yes, 730 N is going to overcome that and actually accelerate this.
The two crates will accelerate in the direction of the force at a rate of ( 730  271.0065 ) / ( 110 + 75 ) = 2.48105 m/s^{2} in Singapore, ( 730  272.64375 ) / ( 110 + 75 ) = 2.47220 m/s^{2} in Helsinki or ( 730  45.0105 ) / ( 110 + 75 ) = 3.70265 on the Moon.
This means that the net horizontal force on the 75 kg crate is 75 * ( 730  271.0065 ) / ( 110 + 75 ) = 186.07845 N in Singapore, 75 * ( 730  272.64375 ) / ( 110 + 75 ) = 185.41470 N in Helsinki or 75 * ( 730  45.0105 ) / ( 110 + 75 ) = 277.69845 N on the Moon. All six of those fifth decimal place figures have been adjusted up because the sixth digit in each case was 5. Freaky, eh?
Likewise, the net horizontal force on the 110 kg crate is 110 * ( 730  271.0065 ) / ( 110 + 75 ) = 272.91505 N in Singapore, 110 * ( 730  272.64375 ) / ( 110 + 75 ) = 271.94155 N in Helsinki or 110 * ( 730  45.0105 ) / ( 110 + 75 ) = 407.29105 N on the Moon.
Quick check: ( 272.91505 / 186.07845 ) * 75 = 110.00000 which is very reassuring.
The horizontal forces on the 110 kg crate are friction and the push it's getting from the 75 kg crate. Friction under the 110 kg crate is 0.15 * 110 * g, which is 161.139 N in Singapore, 162.1125 N in Helsinki or 26.763 N on the Moon. This makes the push from the 75kg crate ( 272.91505 + 161.139 ) = 434.054 N _{( 3 d.p. )} in Singapore, ( 271.94155 + 162.1125 ) = 434.054 N _{( 3 d.p. )} in Helsinki or ( 407.29105 + 26.763 ) = 434.054 N _{( 3 d.p. )} on the Moon.
(Yes, the bunkerbusting bomb already provided this answer but I like posting fifth decimal places, dammit!)
It's usually better to avoid the specific solutions like this and come up with a general one that can be used for other situations without having to go through all the calculations again. As shown above, the local value of g actually cancels itself out and the Singapore / Helsinki / Sea of Tranquility part of it was quite unnecessary!
Total friction R_{t} under the crates is
u * ( M_{1} + M_{2} ) * g
The two crates will accelerate in the direction of the force at a rate of
a = ( F_{t}  R_{t} ) / ( M_{1} + M_{2} )
= ( F_{t}  u * ( M_{1} + M_{2} ) * g ) / ( M_{1} + M_{2} )
= F_{t} / ( M_{1} + M_{2} )  u * ( M_{1} + M_{2} ) * g / ( M_{1} + M_{2} )
= F_{t} / ( M_{1} + M_{2} )  ug
This means that the net horizontal force on the M_{1} kg crate is M_{1} * ( F_{t} / ( M_{1} + M_{2} )  ug )
Likewise, the net horizontal force on the M_{2} kg crate is M_{2} * ( F_{t} / ( M_{1} + M_{2} )  ug )
The horizontal forces on the M_{2} kg crate are friction, R_{2}, and the push it's getting from the M_{1} kg crate, F_{2}.
Friction under the M_{2} kg crate is R_{2} = u * M_{2} * g
This makes the push from the M_{1} kg crate F_{2} = M_{2} * ( F_{t} / ( M_{1} + M_{2} )  ug ) + u * M_{2} * g = M_{2} * ( ( F_{t} / ( M_{1} + M_{2} )  ug ) + ug )
The ((())()(())) is getting ugly but those last two terms do cancel out.
F_{2} = M_{2} * ( ( F_{t} / ( M_{1} + M_{2} ) ) )
F_{2} = M_{2} * F_{t} / ( M_{1} + M_{2} )
... which is M_{2} / ( M_{1} + M_{2} ) * F_{t}, a simple ratio job as durandal said.
the two crates are in facetoface contact, not edgetoface or edgetoedge contact;
the two crates are cuboids;
the two crates are each resting on a face, with neither resting on an edge and the other crate;
the coefficient of friction provided is between crate and surface;
friction is unaffected by velocity (because they didn't specify only initial acceleration);
the horizontal force specified is pushing the 75 kg crate against the 110 kg crate in a direction normal to their faces in contact;
the horizontal force specified is pushing in line with the centres of mass of both crates;
the horizontal force specified is pushing in line with the centres of friction under both crates;
air (or other fluid) resistance is negligible at the speeds involved (well, it does say "at rest" but if they're under treacle this may be an issue);
the two crates are of similar enough size for neither to topple over the other
and
the crates contents are rigid and cannot move around within the crates ...
Thing to check number 1: total friction under the crates is 0.15 * ( 75 + 110 ) * g, which is 271.0065 N in Singapore, 272.64375 N in Helsinki or 45.0105 N on the Moon. Yes, 730 N is going to overcome that and actually accelerate this.
The two crates will accelerate in the direction of the force at a rate of ( 730  271.0065 ) / ( 110 + 75 ) = 2.48105 m/s^{2} in Singapore, ( 730  272.64375 ) / ( 110 + 75 ) = 2.47220 m/s^{2} in Helsinki or ( 730  45.0105 ) / ( 110 + 75 ) = 3.70265 on the Moon.
This means that the net horizontal force on the 75 kg crate is 75 * ( 730  271.0065 ) / ( 110 + 75 ) = 186.07845 N in Singapore, 75 * ( 730  272.64375 ) / ( 110 + 75 ) = 185.41470 N in Helsinki or 75 * ( 730  45.0105 ) / ( 110 + 75 ) = 277.69845 N on the Moon. All six of those fifth decimal place figures have been adjusted up because the sixth digit in each case was 5. Freaky, eh?
Likewise, the net horizontal force on the 110 kg crate is 110 * ( 730  271.0065 ) / ( 110 + 75 ) = 272.91505 N in Singapore, 110 * ( 730  272.64375 ) / ( 110 + 75 ) = 271.94155 N in Helsinki or 110 * ( 730  45.0105 ) / ( 110 + 75 ) = 407.29105 N on the Moon.
Quick check: ( 272.91505 / 186.07845 ) * 75 = 110.00000 which is very reassuring.
The horizontal forces on the 110 kg crate are friction and the push it's getting from the 75 kg crate. Friction under the 110 kg crate is 0.15 * 110 * g, which is 161.139 N in Singapore, 162.1125 N in Helsinki or 26.763 N on the Moon. This makes the push from the 75kg crate ( 272.91505 + 161.139 ) = 434.054 N _{( 3 d.p. )} in Singapore, ( 271.94155 + 162.1125 ) = 434.054 N _{( 3 d.p. )} in Helsinki or ( 407.29105 + 26.763 ) = 434.054 N _{( 3 d.p. )} on the Moon.
(Yes, the bunkerbusting bomb already provided this answer but I like posting fifth decimal places, dammit!)
It's usually better to avoid the specific solutions like this and come up with a general one that can be used for other situations without having to go through all the calculations again. As shown above, the local value of g actually cancels itself out and the Singapore / Helsinki / Sea of Tranquility part of it was quite unnecessary!
Two crates, of mass M_{1} kg and M_{2} kg, are in contact and at rest on a horizontal surface. An F_{t} force is exerted on the M_{1}kg crate. If the coefficient of kinetic friction is u, calculate (a) the acceleration of the system, and (b) the force that each crate exerts on the other.
Total friction R_{t} under the crates is
u * ( M_{1} + M_{2} ) * g
The two crates will accelerate in the direction of the force at a rate of
a = ( F_{t}  R_{t} ) / ( M_{1} + M_{2} )
= ( F_{t}  u * ( M_{1} + M_{2} ) * g ) / ( M_{1} + M_{2} )
= F_{t} / ( M_{1} + M_{2} )  u * ( M_{1} + M_{2} ) * g / ( M_{1} + M_{2} )
= F_{t} / ( M_{1} + M_{2} )  ug
This means that the net horizontal force on the M_{1} kg crate is M_{1} * ( F_{t} / ( M_{1} + M_{2} )  ug )
Likewise, the net horizontal force on the M_{2} kg crate is M_{2} * ( F_{t} / ( M_{1} + M_{2} )  ug )
The horizontal forces on the M_{2} kg crate are friction, R_{2}, and the push it's getting from the M_{1} kg crate, F_{2}.
Friction under the M_{2} kg crate is R_{2} = u * M_{2} * g
This makes the push from the M_{1} kg crate F_{2} = M_{2} * ( F_{t} / ( M_{1} + M_{2} )  ug ) + u * M_{2} * g = M_{2} * ( ( F_{t} / ( M_{1} + M_{2} )  ug ) + ug )
The ((())()(())) is getting ugly but those last two terms do cancel out.
F_{2} = M_{2} * ( ( F_{t} / ( M_{1} + M_{2} ) ) )
F_{2} = M_{2} * F_{t} / ( M_{1} + M_{2} )
... which is M_{2} / ( M_{1} + M_{2} ) * F_{t}, a simple ratio job as durandal said.
Oh, Willie McBride, it was all done in vain.
 Sableagle
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Re: very basic physics question (thats driving me nuts)
My response was lacking, and for this I apologise.
I should have included this assumption:
that the crates are structurally sound and the horizontal force is spread out over the M_{1} crate sufficiently to preserve structural integrity of the crate.
Without that assumption, the force could be on an area of, oh, π( 0.65 * 0.45 * 25.4 / 2 )^{2} mm^{2} and result in a small hole through the crate, up to about 75cm deep.
I should have included this assumption:
that the crates are structurally sound and the horizontal force is spread out over the M_{1} crate sufficiently to preserve structural integrity of the crate.
Without that assumption, the force could be on an area of, oh, π( 0.65 * 0.45 * 25.4 / 2 )^{2} mm^{2} and result in a small hole through the crate, up to about 75cm deep.
Oh, Willie McBride, it was all done in vain.
Re: very basic physics question (thats driving me nuts)
Sableagle wrote:My response was lacking, and for this I apologise.
I should have included this assumption:
That's not the only assumptions you missed. There's the assumption the OP still cares about this problem, that they still visit here, and, heck, that they haven't exited this plane of existence!
To be fair, you weren't the one with the 6y bump though!
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