Data analysis procedure #4: Calculate the position of the third clamp using the second condition for equilibrium and compare with the experimental result by finding the percent difference. 54 Created with Scanner Pro The % difference = Lexpi ched x 100% Kavg The % difference since the force acting on the stick by the fulcrum is unknown, simplify your calculation by calculating the torques about this point. Show work. 5. Now remove the third clamp and slide the half-meter stick so that the knife-edge of the first clamp and the fulcrum are at the 15 cm mark (Figure 3). Then balance the stick by sliding the second clamp with the 100 g hanging mass until equilibrium is obtained. Record the position of the second clamp. *3.C com center point mass of I clamp = cozly mass of steck = 30 40 50 0, 03 560ly Figure 3 100 gData analysis procedure #5: Caleulate the position of the second clamp using the second condition for equilibrium and compare with the experimental result by finding the percent difference. Show work. . Remove the system from the fulerum, attach a hook to the knife-edge in the upward position, and suspend the whole system from a spring balance (Figure 4). This will give you the upward force provided by the fulerum, Data analysis procedure #6 Calculate the upward force provided by the fulcrum using the first condition for equilibrium and compare with the experimental result by finding the percent difference. Show work. 7. Place a clamp with a hanging mass of 300 g at the 30.0 cm mark of the half-meter stick (Figure 5). Place the second clamp at the 5.0 cm mark on the fulcrum and the third clamp at the 40.0 55 Created with Scanner Pro cm mark with a hook upward. Hang the upward hook to a spring balance and find the reading on the balance. Be sure to keep the half-meter stick horizontal. 10 20 300 g Figure 5Data analysis procedure #7: Calculate the upward force provided by the spring balance using the second cnlnditiun for _ equilibrium and compare with the experimental result by finding the percent difference. Shiw work. 8. Interchange the spring balance with the fulcrum and read the spring balance. Data analysis procedure #8: Check the sum of the two spring balance readings from procedures 7 and 8 against the total weight to verify the first condition for equilibrium, Show work. Questions: 1. Describe what the motion of a rigid system will be if: a) The first condition for equilibrium is not satisfied. b) The second condition for equilibrium is not satisfied. ) Meither condition for equilibrium is satisfied. 2. If the numbers of clamps on each side of the support fulcrum are the same, say one and one, can their weight be omitted from the torque caleulation? Explain. 3. Ifthree forces act on an extended object and the object is in equilibrium, prove that the lines of | action of these three forces intersect at one point. ' 4. Use two fingers to support a meter stick, one at the 10 cm mark and another at the 65 cm mark, | Move the two fingers closer slowly until they meet. Where do the two fingers meet? Repeat with different starting positions for the two fingers and explain the results. when an object is in equilibrium the net force acting on that object has to be zero. This is the first condition for equilibrium. > F=0 [n the two-dimensional case, it is equivalent to two-component equations: L0, s, B0 The rotational etfect of a force acting on an extended object depends on three factors: the ma gnitude of the force, the direction of the force, and the location of the axis of rotation. The torque T is defined as T=Fr where r is the moment anm which is the perpendicular distance from the axis of rotation to the line of action of the force. When the line of action of the force passes through the axis of rotation, the torque of the force aboul this axis is zero. This can be used to simplify our calculation when we choose the axis of rotation on the line of action of an unknown force. The torque of the unknown force will be zero. Ifan extended object is in equilibrium, then the net torque of the forces acting on the object about any axis of rotation should also be zero. This is the second condition for equilibrium: ZT {about any axis of rotahon) =0 In this formula, we take counterclockwise torque to be positive and clockwise torque to be negative, The above equation can also be wrillen as Z (clockwise) = Z (counterclockwise)