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Download Equilibrium is not just translational, is is also rotational. While a set
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Equilibrium is not just translational, is is also rotational. While a set of forces may not change the velocity of an object, it may change its speed of rotation. The lever arm is the distance between the line of action (direction of force applied) and the axis of rotation. Torque Ƭ is the magnitude of the force multiplied by the lever arm. Ƭ = Fl The torque is positive if the force tends to produce a counterclockwise rotation, and negative if it tends to produce a clockwise rotation. The unit is the newton•meter (N•m). Ex. 1 - A force whose magnitude is 55 N is applied to a door. However, the lever arms are different lengths in each of three cases: (a) l = 0.80 m, (b) l = 0.60 m, (c) l = 0. Find the magnitude of the torque in each case. Ex. 2 - In your ankle joint, your Achilles tendon is attached to the heel. If the tendon exerts a force of F = 720 N at a n angle of 55° to the lever arm, find the torque of this force about the ankle joint, which is 3.6 x 10-2 m from the point of attachment. If a rigid body is in equilibrium its motion does not change. Therefore, the net force is zero. For twodimensional motion, the sum of the x and y components of the forces are both zero. In these conditions, we exclude all internal forces, because these are all action-reaction pairs. Under equilibrium conditions, there is also no net torque. If there is a net torque, the object will undergo angular acceleration. If an object is in equilibrium, there is no translational acceleration and no rotational acceleration. ∑Fx = 0 and ∑Fy = 0 ∑Ƭ = 0 The above must be true for all equilibrium conditions. Ex. 3 - A woman whose weight is 530 N is poised at the right end of a diving board, whose length is 3.90 m. The board has negligible weight and is bolted down at the left end, while being supported 1.40 m away by a fulcrum. Find the forces F1 and F2 that the bolt and the fulcrum, respectively, exert on the board. The location of the axis is completely arbitrary, because if an object is in equilibrium, it is in equilibrium with respect to any axis whatsoever. Ex. 4 - An 8.00-m ladder of weight WL = 355 N leans against a smooth vertical wall. The wall exerts only a normal force and no frictional force. A firefighter, whose weight is WF = 875 N, stands 6.30 m from the bottom of the ladder. Assume that the weight of the ladder acts at the ladder’s center and neglect the weight of the hose?!?! (Leggs, maybe?) Find the forces that the wall and ground exert on the ladder. Choosing the direction of an unknown force backward in the free-body diagram simply means that the value determined for the force will be a negative number. Ex. 4 - A body builder, strengthening his shoulder muscles, holds a dumbbell of weight Wd. his arm is extended horizontally and weighs Wa = 31.0 N. The deltoid muscle is assumed to be the only muscle acting. The maximum force M that the deltoid muscle can supply is 1840 N. What is the weight of the heaviest dumbbell that can be held, and what are the horizontal and vertical force components, Sx and Sy , that the shoulder joint applies to the left end of the arm?