Hey, I was doing some statics problems involving friction for fun. I came across this one problem that concerned me greatly. The free body diagram looks something like what is posted. I know the picture is not great but I have the solution for this problem and I know the correct numbers, so you will have to trust me on that part. If u work the problem but get a different anwser it is probably because my picture is bad. To start off, there is a 300N applied force, which you can see on the left. This acts on a chain, which is represented by the THIN black lines. They are at a 60 degree angle. If you sum the forces where the three pieces of chain are, you can determine the tension in each of the two angled chains. Next what the solution did, was take the moments about the hinge of the tongs at point c for ONE of the tong members. Then you have a nomral force 75 mm away, the tension force acting both 75mm, and 50mm away (when done component wise), and a friction force acting 50mm away. Now comes the constraint that if this is impending motion, that the friction force is equal to F=mu*norm, and mu=0.5. If you plug this in, you only have one variable now, the normal, and you can solve it to find the friction force as well. But the friction force you get is equal to 107 on each side of the tong, or 214N of friction. But this is NOT equal and opposite to the applied 300N! So how do we account for the lack of sum of forces equal zero in the x direction? If we look at the system as being the tongs and the chain, then we have a 300N force at one end, and a 214N friction force opposite at the other end. Its NOT equal to zero, yet it is static.
Right. Given the chain angles, you can find the tension in the chains. Right. This should completely determine the normal force acting on the tong end. (The friction will not exert a torque.) I'm not clear on what the problem is asking you to find. Once you have the normal force, you can find the coefficient of friction that will just allow equilibrium. (Assuming that's what you are supposed to find.) On the other hand, if the coefficient is given as 0.5 (as you state later on), then the problem seems overdetermined. [itex]\mu N[/itex] will be the maximum static friction; the actual friction force will be less. (But that violates your statement that it is just about to move.) I'm not clear from your description exactly what you are supposed to calculate. What does the problem ask for? Sorry, but I don't understand this part. If you are summing forces on just one tong, you must include the force from the other tong at the hinge. But that's not needed. What am I missing? What were you supposed to find?
The forces exerted at the hinge are internal forces; they cannot balance the 300N external force. That 300N force must be balanced by the frictional force, the only other horizontal force acting on the tongs.
no the hinge does exert a linear force it gives no torque. if you try to assemble the system and fail to anchor the hinge to something that provides that force the system as described will accelerate to the left of the diagram bow to my superior knowledge
ok guys let me clear this stuff up for yall. The friction force on EACH tong is 107N. There are two parts of the tong that clamp or bite. So the total friction is twice this amount, 214N. So in effect what you have is 300N pulling to the left, but only 214N of friction to the right! You cant have more friction because the coefficient of static friction is given as 0.5. The hing is NOT anchored, and is ALLOWED to move. What I did not mention, as it is not imporant to my question, is that the tongs are pulling a heavy crate on a floor with friction. The reason this second detail is not important is because the crate sees a pulling force of only 214N, not 300. So even in that closed system, you have 300N pulling and 214N resisting. As Doc correctly stated, the pin is an INTERNAL force to the system. I still dont see where the remainder of that 300N went! ( I will upload a picture of the exact solution for you guys once I get home, im at work right now.)
I suspect you forgot to take into account one component of the force the chain exerts on one tong 'handle'. The tension in both chains attached to the tongs is 300 N (because of the 1-2-sqrt(3) triangles). For one tong, the horizontal component of that force is 150 N, and the vertical component is 150*sqrt(3) N (I'm not paying attention to the correct signs here, the setup is simple enough to do the calculation 'by inspection'). I gather from the figure that the vertical distance from the point where the chain is attached to the hinge is 50 mm, and the horizontal distance is 75 mm. The total moment exerted by the chain on the tong, measured from the hinge, is: [tex]M=150\cdot \sqrt{3} \cdot 75 + 150 \cdot 50[/tex] Nmm. To find the normal force the gripping end exerts on the block thing, divide this moment by 75 (the horizontal distance from the hinge to the gripping point) to get: [tex]F_{N} = 150 \cdot \sqrt{3} + 100 = 359.8[/tex] N, approximately. The maximum lateral force that can be applied before the grip starts to slip is half of that, which is about 160 N, and the total maximum force that can hence be applied on the left is about 320 N. No worries! We're still safe. Is 20 N an acceptable margin to consider that the whole thing is 'just about to slip'? P.S. substitute 'torque' for 'moment' everywhere if you think my terminology is a bit funny.
If you look at the tonges, each tong is in equilibrium when you consider the a. tension, b. pin reactions at c, c. the normal and friction forces at the tong lip. Thats ok thats fine and dandy. But consider both tongs, and the chain, as one system. then the only forces are the 300N force, and the two friction forces and normal forces acting on the new free body diagram. The normal forces obviously cancel due to symmetry. But the friction force and the applied load P do not cancel! See what I mean now?
the analysis is false, the tongs will slip and accelerate to the left no matter what force is applied the design of the tongs needs to be fixed by changing the angle of the chain. consider the system if you lengthened the chains so that the angle is close to 90 then it is obvious that the clamp will slip off. If you shorten the joining chain so that the angle is less this will cause the grip to be better for less applied force. It would be a good question to find the minimum angle that the chain could be at in orderto grip. The question is wrong PS appologee to Doc, I was not trying being rude
Yeah, I used the wrong geometry in my calculation where I was following the initial description, as it turned out. I blame cyrusabdollahi's drawing skills! ;) I'm inclined to agree with GNPatterson's conclusion: the total frictional force of the tongs on the block is always less than the tension force applied to the left - because if this is the cae in one situation it's the case for all situations, due to simple scaling of all forces.
Yes. Now it's quite clear. And your concern is justified. The problem is bogus. It is based on the assumption that a towing force of 300N is applied, but that is physically unrealizable. (In fact, given the angles and coefficients of friction, this arrangement seems unable to support any towing force!) Can you please tell us the title and author of the book where you found this?
take a guess doc al, AHAHAHAHA HIBBELERRRRRRRRRRRR, that guy has so many bad problem in his book it is really quite INSANE! Im glad to hear you agree with me, for a moment there I was thinking that I dident learn a thing when I took statics! Could you explain please? I mean, each of his steps does make good logical sense though.
The steps make logical sense only up to the point where they contradict the laws of physics! He starts with the bogus assumption that P = 300N and never looks back. He arrives at a friction force insufficient to balance the stipulated force P: a contradiction. To understand my statement ("In fact, given the angles and coefficients of friction, this arrangement seems unable to support any towing force!"), just work the problem backwards: Find a value of P that satifies the conditions of the problem. We know that: [itex]P = 2F[/itex] [itex]2T \sin (60) = P[/itex] [itex]F = \mu N[/itex] Now, using the above, rewrite the torque equation completely in terms of P. What value of P is required to satisfy that equation?
I gotcha! EXACTAMUNDO! Yeah thats not a real word, sue me! At least I was paying attention when doing the problem, for a change! Belive it or not Ive done every single problem in the book, (almost). I still have 3 more chapters to go, then ill be satisfied that i understand statics well enough to go back and start reviewing dynamics! Its amazing how much more sense it makes when you have time to sit back and THINK and not CRAM a chapter a week! I hate the college learning style.....grrrr! But to be realistic, it would take 20 years to graduate at my pace! Im so slow sometimes :-)