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Transcript
AP Physics – Torque Forces: We’ve learned that forces change the velocity of an object. But what does it take to change the angular velocity of a thing? Well, forces are involved, but the force has to be applied in a special way. We call this special applied force a torque. There are many ways to apply a force to a system that can rotate. In the drawing below we have a turntable that can spin. If we just push sideways on the thing, as in the drawing to the left, we will not make it spin. We basically would be trying to tip it over. But if we apply a force tangent to the disc as in the drawing to the right, it will spin. This force is perpendicular to a radius of the circular path. A force that is applied perpendicular to the circular path at some distance from the spin axis is called a torque. Torques change angular velocity. The symbol for torque is the Greek letter . Torque is given by this equation: rF sin r is the distance to the center of spin from where the force is applied. This variable is often called the lever arm. F sin be the force component that is perpendicular to the lever arm. F sin F r If the angle is 90, then the force is perpendicular to the lever arm, the sine is one, and the equation for torque is simply: Fr Note for some unknown reason, the force is written first and then the lever arm in this equation. 197 You can see that the unit for torque is going to be a newton meter (nm). We leave it like that. This looks very similar to the unit for work, the joule, but it is quite different. So energy and work are in joules and torque is left in newton meters. Torque is a vector quantity. 125 N is applied to a nut by a wrench. The length of the wrench is 0.300 m. What is the torque? Fr 125 N 0.300 m A torque of 857 Nm is applied to flywheel that has a radius of 45.5 cm. What is the applied force? Fr 37.5 Nm F r 1 857 Nm 0.455 m 1880 N You push on the door as shown in the drawing. What is the torque? rF sin 330 N 330 N 1.5 m sin55.0 1.5 m 410 Nm Multiple Torques: 55.0 What happens if two or more torques act on an object at the same time? Two forces are applied to the object in the drawing to the right. The object is free to rotate about the spin axis. Both cause a torque. F1 causes a CCW (counter clockwise) rotation around the axis. F1 spin axis r1 r2 F2 causes a CW (clockwise) rotation around the axis. If a torque causes a clockwise rotation, it is positive. If a torque causes a counter clockwise rotation, it is negative. F2 198 The sum of the two torques would be: =1 2 F2r2 F1r1 Equilibrium and Torque: If an object is in angular equilibrium (sometimes called rotational equilibrium), then it is either at rest or else it is rotating with a constant angular velocity: If object is in rotational equilibrium, the net torque about any axis is zero. This means that the sum of the torques acting on the object must be zero. =0 Static equilibrium exists when an object has no motion, either linear or angular. There are two conditions which must exist in order to have your good old static equilibrium: The net force must be zero and the net torque must be zero. F=0 =0 This gives us some very powerful tools to solve static problems. We can analyze a system and look at the forces acting on it, and we can also look at the torques that act on it. We’ll be able to do some really cool stuff. 199 Two metal orbs are attached to a very lightweight rigid wire. They are suspended from a rigid point on the overhead as shown. The system does not move. Calculate the distance from the suspension line to the center of gravity on the right sphere. Since the system is at rest, the sum of the torques and the sum of the forces must be zero. Let’s look at a FBD and a drawing showing the two torques: 4.0 kg 1.0 kg F 45.0 cm m1 g r1 m2 g Without using the torque equilibrium, we could not solve the problem. The sum of forces would simply tell us that the upward force would be equal to weight of the two balls. ? r2 m1 gr1 m2 gr2 Using torque, however, allows us to solve the problem. All we have to do is add up d’ torques: 1 2 0 r2 m1gr1 m2 gr2 0 1.0 kg 0.45 m 4.0 kg m2 gr2 m1gr1 r2 m1 gr1 m2 g 0.11 m or 11 cm Torque problems, as you have just seen, are fairly simple. 200 Now we’ll do a classic teeter-totter beam problem. A teeter-totter is in equilibrium as shown. The block on the left has a weight of 625 N. The beam itself has a weight of 32.5 N. What is the mass of the second block? 625 N ? This is a pretty simple problem, we can solve it using the torques. The sum of the torques must be zero: =0 There are three torqes, 1 (from the 625 N rock) and 2 from the other rock. The weight of the beam (Fbeam), even though it has a significant amount of weight, does not cause a torque because the weight acts at the CG of the beam which is also the center of rotation. Thus the lever arm is zero. 1 2 0 1.10 m F1 3.30 m F2 Fbeam F1r1 F2r2 0 2 is positive (CW) and 1 is negative (CCW) F1r1 F2r2 F2 F1r1 r2 685 N 1.10 m 3.30 m 228.3 N To find the mass we use the second law: F ma m F a kg m 1 228.3 s 2 9.8 m s2 23.3 kg Another similar problem. A 50.0 N seesaw supports two people who weigh 455 N and 525 N. The fulcrum is under the CG of the board. The 525 N person is 1.50 m from the center. (a) Find the upward force n exerted by fulcrum on the board. (b) Where does the smaller person sit so the seesaw is balanced? 1.50 m x 201 First, let’s draw a FBD n r2 r1 We know that the system is in static equilibrium, so we can analyze the forces. In the y direction, the sum of the forces must be zero. Fy 0 (a) F1 and F2 are the weight of the two men, FT is the weight of the teeter-totter, and n is the normal force. FT F1 F2 We can write this out: n F1 F2 FT 0 Now we can solve for the normal force, this is the upward force exerted on the board by the support stand. n F1 F2 FT 0 525 N 455 N 50.0 N 1 020 N (b) To find the distance the second man must be positioned from the center, we must analyze the torques. y 0 r2 F1r1 F2 1 2 0 525 N 1.50 m 455 N F1r1 F2r2 0 F1r1 F2r2 1.73 N 202 AP Physics – Applying Torque It is now time to go after some problems that are more complicated. You will find these to be a lot of phun. Honest. A uniform beam is supported by a stout piece of line as shown. The beam weighs 175 N. The cable makes an angle of 75.0 as shown. Find (a) the tension in the cable and (b) the force exerted on the end of the beam by the wall. 75.0 4.00 m We can solve this problem by summing forces and adding up torques. T R 75.0 First let’s draw a FBD: We have three forces acting on the beam. The weight of the beam which acts at the center of the beam (its CG), FB. F B The tension in the cable, T. And the force exerted by the wall on the beam, R. (The wall is pushing the beam up and out.) Think about this as an application of Newton’s Third Law. R must be acting in that way to counteract the Tension in the cable. (a) Let us first sum the torques. The pivot point is the end of beam where it meets the wall. Therfore R exerts no torque as its lever arm is zero. We only have two torques to deal with and, of course, they add up to zero. Torque one is exerted by the tension in the cable and torque two is caused by the weight of the beam. The force for this torque is applied at the CG, which is at the center of the beam. Only the vertical component of the tension causes its torque so: cable B 0 T FB r 2 r sin r Tr sin FB 0 2 175 N 2 sin 75.0o 90.6 N (b) Next we can sum up the forces: 203 T cos RX 0 x direction: y direction: We can solve the x direction equation for RX: 90.6 N cos75.0 RX T cos T sin FB RY 0 23.4 N Next we solve for RY: RY FB T sin 175 N 90.6 N sin75.0o 87.5 N We’ve found the x and y components for R, so now we can find the magnitude of the vector using the Pythagorean theorem: R Ry 2 Rx 2 87.5 N 2 23.4 N 2 90.6 N A beam is supported as shown. The beam is uniform and weighs 300.0 N and is 5.00 m long. A 635 N person stands 1.50 m from the building. (a) What is the tension in the cable and (b) the force exerted on the beam by the building? We draw a FBD. ALWAYS DRAW THE FBD!!!!! R T 55.0 55.0 5.50 m (a) Sum of torques: beam man cable 0 FB Fm rm FB rB TrC sin 0 Fm T 635 N 1.50 m 300.0 N 2.50 m sin 55.0o 5.00 m T Fm rm FB rB sin rC 416 N (b) We can resolve R and T into components and then sum the forces in the x and y direction. All forces must add up to equal zero. 204 RX T cos 0 Ry T sin FB Fm 0 RX T cos 0 RX T cos Ry T sin FB Fm 0 416 N cos55.00 Ry FB Fm T sin Ry 300.0 N 635 N 416 N sin R Ry 2 Rx 2 238 N 624 N 624 N 2 238 N 2 668 N Fabulous Ladder Problems: Ladder problems are very popular. The basic idea is that you have a ladder leaning against a wall (which is usually frictionless). The ladder is held in place by the friction between its base and the deck it rests upon. We’re given the situation and then required to figure out various things – the angle the ladder makes with the deck, the friction force, the coefficient of friction, the force exerted on the top of the ladder by the wall, &tc. Let’s go ahead and do a simple problem. A uniform 250.0 N ladder that is 10.0 m long rests against a frictionless wall at an angle of 58.0, the ladder just keeps from slipping. (a) What are the forces acting on the bottom of the ladder? (b) What is the coefficient of friction of the bottom of the ladder with the ground? 10.0 m Draw a FBD. The forces acting on the ladder are: the weight of the ladder FL, the frictional force f, The force the deck pushes up on the ladder with F1, and the force exerted by the wall on the top of the ladder F2. 58.0 Now we look at the forces acting on ladder - they have to add up to zero. F2 Fy 0 F1 FL 0 Fx 0 f F2 0 F1 FL 250 N F1 f F2 We have to find either F2 or else f. But we need more info, don’t we? You bet we do. Blessed by good fortune as we are, we instantly recognize that we can make use of the torque equilibrium deal. First we make a drawing showing all the torques acting on the ladder. (Actually we’re only looking at the forces that are perpendicular to the lever arm.) FL f 205 The pivot point is the base of the ladder. Neither the friction or F1 cause a torque as their lever arm is zero. F2 The weight of the ladder causes a CCW torque. The lever arm from F2 causes a CW torque. 12.0 m The torques add up to zero. 0 6.00 m ladder wall 0 FL The angle is, using geometry clearly going to be: 90o 90o 58o F2 32o F2 cos d 2 FL cos d1 0 The torques are: FL cos d1 cos d 2 250.0 N cos 58.0 6.00 m cos 32.0 Solve for F2: 78.1 N 12.0 m The frictional force (the other force acting at the base of the ladder is therefore: f 78.1 N (b) Find the coefficient of friction: f n f n 78.1 N 250.0 N 0.312 Whew! Can we make it worse? You bet. 206 A 15 m, 500.0 N uniform ladder rests against a frictionless wall. It makes 60.0 angle with the horizontal. Find (a) the horizontal and vertical forces on the base of the ladder if an 800.0 N fire fighter is standing 4.0 m from the bottom. If the ladder is on the verge of slipping when the fire fighter is 9.0 m from the bottom of the ladder, (b) what is the coefficient of static friction on the bottom? 15 m FY 0 F1 FL FF 0 4.0 m F1 FL FF 60.0 F1 500 N 800 N F1 1300 N F2 FX 0 F2 F2 f 0 f F2 F1 FL Let’s look at torque to find F2: 0 Pivot point is at the base of the ladder: 90o 90o 60o 30o 2 L F 0 FL f F2 cos r2 FL cos rL FF rF 0 F2 F2 FF FL cos rL FF rF cos r2 FF 500.0 N cos60.0 7.50 m 800.0 N cos60.0 4.00 m F2 270 N cos30.0 15.0 m so f 270 N 207 So the force up is F1 1300 N The horizontal frictional force is: f 270 N (b) When the fire man is at 9.0 m (we’ll figure this from the bottom of the ladder), then We can use the same equation as we used to find F2 since the only thing that has changed is the distance of the firefighter from the bottom of the ladder. F2 F2 FL cos rL FF rF cos r2 F2 500.0 N cos 60.0 7.50 m 800.0 N cos 60.0 9.00 m cos 30.0 15.0 m F1 f 421 N We can now find the coefficient of static friction for the bottom of the ladder and the deck. f s n s f n 421 N 1300 N 0.32 FF FL f All there is to it. 208