kingwinner Posted April 4, 2006 Report Share Posted April 4, 2006 The following are 2 of the Sir Isaac Newton Contest questions in the past years that I found in my textbook. Unfortunely, I am having a fair amount of problems and questions with these. It would be nice if somebody can explain, I would appreciate! :rolleyes: 1) First, I have assumed that the applied force on m1 is not equal to the force of m1 on m2, I then got the force of m1 on m2 (or the contact force) is 1.0N, is this correct? (there is no answer in the textbook) Second, I would like to ask why wouldn't the applied force on m1 be equal to the force of m1 on m2? Shouldn't m1 be just acting as a mass transferring the 3.0N applied force directly to the next box? I don't get the idea... 2) This question I am having a lot of troubles with! Is the string fixed to the poles on both ends? But it seems that there are pulleys around, is the string fixed to the pole below the pulley? There is only 1 string in this system, so the tension force must be the same throughout the length of the string. So there are 2 tension forces pulling in opposite directions along the string at point C, but since the angle aren't equal, which, with the force of gravity, should balance out to zero. However, the horizontal forces don't balance out, this is where I got stuck... And finally, what does minimum breaking strength actually mean? I am not sure what the question is looking for! Quote Link to comment Share on other sites More sharing options...
Pyrotex Posted April 6, 2006 Report Share Posted April 6, 2006 ...First, I have assumed that the applied force on m1 is not equal to the force of m1 on m2, I then got the force of m1 on m2 (or the contact force) is 1.0N, is this correct? (there is no answer in the textbook)Let's assume the applied force on m1 by m2 (and on m2 by m1, for they ARE equal) is a million N. Obviously wrong, but why? How the hell could you be applying 3.0 N to the first block if its force on the second is so huge? If a million N is obviously too big, what is the answer? 3.0 N. Same as your force on the first block. ......This question I am having a lot of troubles with! Is the string fixed to the poles on both ends? But it seems that there are pulleys around, is the string fixed to the pole below the pulley?...And finally, what does minimum breaking strength actually mean? I am not sure what the question is looking for!If you tied the string to the chicken and pulled it up, letting the chicken just hang from the string, then the minimum breaking strength (mbs) of the string must be equal to the weight of the chicken -- the mbs/string would be 2.0 N. No, the tension in both arms of the string may not be the same. Ignore the pulleys -- the string is essentially fixed at each end. I would solve it by taking the vertical and horizontal components of the forces.Fv (of chicken's weight) = Facv + Fbcv (sum of the vertical comps in strings)or Fac (sin 30) + Fbc (sin 45) = 2 kg * G = 2 N .5 Fac + .71 Fbc = 2 The chicken isn't moving, so Fach = Fbch (rather than Fac = Fbc !!!!! This is the sticking point!!!)or Fac (cos 30) = Fbc (sin 45) .87 Fac = .71 Fbc We have Fac + 1.41 Fbc = 4 Fac = 4 - 1.41 Fbcand Fac = .71/.87 Fbc = .82 Fbc substituting4 - 1.41 Fbc = .82 Fbc0.6 Fbc = 4Fbc = 6.7 N Notice this is way more than the weight of bird. Fac = .82 Fbc = 5.4 N The mbs/string must be 6.7 N, or otherwise the short stretch would break. Now, convince yourself that the tension in the two parts of the string do NOT have to be equal. Do this by exagerating the problem. Try angles of 1 degree and 89 degrees. Remember, the chicken is holding onto the string, and not moving -- ie, in equilibrium.. Quote Link to comment Share on other sites More sharing options...
Qfwfq Posted April 7, 2006 Report Share Posted April 7, 2006 If a million N is obviously too big, what is the answer? 3.0 N. Same as your force on the first block.Now that deserves a good whack on the knuckles! It says frictionles table so they get accelerated. If you want to help King with it, think more carefully and remember your freshman course exercises. Quote Link to comment Share on other sites More sharing options...
sigsfried Posted April 7, 2006 Report Share Posted April 7, 2006 Q1. The total mass is 3kgThe acceleration in 1m/sThe force required on the second one is therefore 1N. Otherwise the second block would accelerate at a different speed to the second one. Quote Link to comment Share on other sites More sharing options...
Qfwfq Posted April 7, 2006 Report Share Posted April 7, 2006 Quite right. The two accelerations couldn't be different, in steady force regime. Quote Link to comment Share on other sites More sharing options...
sigsfried Posted April 7, 2006 Report Share Posted April 7, 2006 Quite right. The two accelerations couldn't be different, in steady force regime. Wouldn't matter is it wasn't a steady force. If the two accelerations are different one of two unphysical situations occur. Bock one passes through block two or block two seperates in which case how is it still accelerating. This is true unless block 1 is slowed. Quote Link to comment Share on other sites More sharing options...
Qfwfq Posted April 7, 2006 Report Share Posted April 7, 2006 :shrug: A knock can propagate through solid objects in contact, with the last one getting projected away. The acceleration lasts briefly. This scenario, called impulsive regime, requires a duration comparable to the time it takes to propagate through the solid object. Quote Link to comment Share on other sites More sharing options...
Pyrotex Posted April 7, 2006 Report Share Posted April 7, 2006 Now that deserves a good whack on the knuckles!It says frictionless table ...:cocktail: :cocktail: :) :) :doh: "frictionless" yes Quote Link to comment Share on other sites More sharing options...
Qfwfq Posted April 7, 2006 Report Share Posted April 7, 2006 Now, for homework :cocktail: work it out supposing a coefficient of friction of 0.16 for initiating motion. Quote Link to comment Share on other sites More sharing options...
sigsfried Posted April 7, 2006 Report Share Posted April 7, 2006 I got 0.84 N That can't be right I'll go away and try again Quote Link to comment Share on other sites More sharing options...
sigsfried Posted April 7, 2006 Report Share Posted April 7, 2006 1 N. It is independant of co-efficient of friction Quote Link to comment Share on other sites More sharing options...
Qfwfq Posted April 7, 2006 Report Share Posted April 7, 2006 :cocktail: Quote Link to comment Share on other sites More sharing options...
sigsfried Posted April 7, 2006 Report Share Posted April 7, 2006 Not convinced? Let coefficent = a Net force on body = 3-3a=3(1-a) (Body being combined system)Acceleration of body=3(1-a)/3=1-aNet Force required to produce this acceleration on mass 2=1-aForce due to Fricition=aTHerefore force accelertaiong body=1-a+a=1 Quote Link to comment Share on other sites More sharing options...
Qfwfq Posted April 10, 2006 Report Share Posted April 10, 2006 Not convinced?Not at all! Let coefficent = a Net force on body = 3-3a=...Why the 3a? What are you multiplying a by? Quote Link to comment Share on other sites More sharing options...
sigsfried Posted April 10, 2006 Report Share Posted April 10, 2006 Not at all! Why the 3a? What are you multiplying a by?THe weight of the system. As frictrion equals coefficient of Friction times weight Quote Link to comment Share on other sites More sharing options...
Qfwfq Posted April 10, 2006 Report Share Posted April 10, 2006 In which unit? The friction coefficient is usually defined with transverse force and weight in the same units. Quote Link to comment Share on other sites More sharing options...
sigsfried Posted April 11, 2006 Report Share Posted April 11, 2006 Fr=uR (friction equals coefficient of Friction times Normal reaction.) Is independant of the units of either Friction or Normal reaction but they must have teh same unit. The coefficient of Friciton is dimensionless so has no units. Quote Link to comment Share on other sites More sharing options...
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