July 21 - Equilibrium Of Non-Parallel Forces 1 Equilibrium Of Non-Parallel Forces INTRODUCTION A particle is in static equilibrium, i.e. will not be accelerated, if the net force acting on it is zero. Another way of expressing this condition is to say that the vector sum of all the forces Fi is zero.1 1nii==∑F0 (1) The same holds true for an extended body provided the lines along which the various forces act all intersect at a common point. If the lines of force do not intersect at a common point the body will be set into rotation even though the vector sum of the forces may equal zero. If we want the term equilibrium to include the absence of a rotational acceleration we must supplement Eq. 1 with the condition that all the torques acting on the body add up to zero as well. What is a torque? Imagine that a force F is applied to various points on a body that is free to rotate around a pivot point P, as shown in Fig. 1. If the line along which this force acts passes through P, no rotation will result. If the force acts at point A1 that lies on a line that passes at finite distance r1 from P, the body will begin to rotate around P. Experience shows that the body will tend to rotate more readily if the same force F is given more leverage, e.g. by applying it at A2, at a distance , from P. In this example the force acts at points P, A12rr >1, and A2, that all lie on a line that is perpendicular to the force. In the general case, shown in Fig. 2, this need not be so. Here the force F acts at a point A that is at a distance r from the pivot point P. Now imagine that this force were transmitted by a string and that a thumbtack were pushed through that string at the point B. Clearly the initial situation would remain quite unchanged2; in other words, the only thing that counts is the perpendicular distance r⊥ from the line of action (force) to the pivot point. This distance is called the moment arm of the force around P. The product of the force F and its moment arm r⊥ is called the torque and is denoted by the Greek letter τ (pronounced tau): +Pr1r2A2A1FFF Fig. 1 1 For a review of vectors, refer to Appendix E: Vectors. 2 As the body starts to rotate, the angle θ will, of course, change differently in the two situations. Physics 635 University of Virginia Summer 2005July 21 - Equilibrium Of Non-Parallel Forces 2 Fr⊥=τ. (2) F+θPBAr⊥r. In the SI system a torque is measured in Newton-meters. A look at Fig. 2 shows that θsinrr =⊥. (3) so that a useful expression of the torque due to a force acting on an arbitrary point becomes θτsinrF= . (4) Fig. 2 Using vector notation, we have . (5) =×τ rF The greater the torque applied to an object, the greater will be the angular acceleration of that object, i.e. the faster it will spin up. The condition for a body to be in rotational equilibrium is that the (vector) sum of the torques acting upon it about any point is zero: ii=∑τ 0 . (6) If the sum of the torques on a body is zero about any one point, it is zero about all points. Hence the location of the point about which torques are calculated in an equilibrium problem is arbitrary. In this experiment you will study the conditions required for a body to be in equilibrium under the action of several forces that are not parallel although they will all lie in one plane, i.e. the torques will be parallel. APPARATUS One drawing board, three steel balls, four clamp pulleys, set of hooked weights, string, thumbtacks, protractor, and washer. WHAT TO DO We have seen that, in order for an extended rigid body to be in equilibrium, the forces acting on it must satisfy two conditions. The sum of the forces must be zero, and the sum of the torques produced by the forces must be zero. For parallel forces the rules for the addition of forces are simple: forces are either positive or negative and one merely takes their algebraic sum. When the forces are not parallel, the rules for addition become more complicated. Forces are no longer merely positive or negative. To deal with this situation, a branch of mathematics, called vector algebra, has been developed (see Appendix E). Similarly, torques are complicated to add when the forces are not coplanar. In this experiment, however, we will restrict ourselves to coplanar forces and so the torques will only be clockwise or counterclockwise (negative or positive). Physics 635 University of Virginia Summer 2005July 21 - Equilibrium Of Non-Parallel Forces 3 1. Since torques complicate matters, we first study a situation without them: Attach three strings to the washer and let them run over the three pulleys, mounted two on one side of the table, and one on the other as shown in Fig. 3. Hook masses, each greater than 200 g, to the ends of these strings so that the washer remains in equilibrium somewhere near the center of the drawing board; this will require some experimentation. Make sure that • the pulleys are oriented so that the strings run through the middle of the grooves, and Fig. 3 • the tops of the pulleys are at approximately the same height as the washer. A large piece of paper tacked to the drawing board under the washer provides a convenient means to draw a diagram of the forces. Clearly the forces are in line with the strings, and can be represented by arrows pointing along the string from the ‘point’ (the center of the washer). Represent the magnitude of the force (Fi = mig) by the length of the arrow. Choose a convenient scale like one centimeter per 10 grams of mass on the string. The resulting diagram should look similar to 4. 2. Since the washer is in equilibrium, we know the vector sum of the three forces is zero. Choose one of the force vectors to be a reference (F3 in Fig. 4) and measure the angles that the other force vectors make with the reference. Using the magnitudes and angles, calculate the parallel and orthogonal components of the force vectors relative to the reference. θ2 θ1 F3 F1 F2 Fig. 4 Question 1: Do the components algebraically add to zero? 3. Of course, you may not get exactly zero because of experimental errors. Estimate how big these errors might be by experimenting: Move the washer 10-20 cm from its equilibrium position, release it, and mark where it comes to a stop; repeat this several times so that you have