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Examples from Chapter 4 Problem 4.31 Two horses pull horizontally on ropes attached to a stump. The two forces, F1 and F2, that they apply to the stump are such that the net (resultant) force R has a magnitude equal to F1 and makes an angle of 900 with respect to F1. Let F1=1300 N (and R=1300 N also). Find the magnitude and direction of F2. Step 1 Draw It! R F2 q F1 Step 2 Break Forces into Components R F2 F2 sin (1800-q) q F1 F2 sin (1800-q)=F2 sin (q) F2 cos (1800-q)=-F2 cos (q) Step 3 Sum the forces in the vertical and horizontal directions, then set them equal to their respective resultant components In the horizontal direction: F1-F2cos(q)=0 In the vertical direction: F2sin(q)=R=1300 N So F2cos(q)=F1=1300 N Thus, q=tan-1(1300/1300)=450 F2=sqrt(2)*1300=1838 N Problem 4.49 The two blocks are connected by a heavy uniform rope with a mass of 4 kg. An upward force of 200 N is applied as shown. A) Draw three free body diagrams: one for the 6 kg block, one for the 4 kg rope, and another for the 5 kg block. For each force, indicate what body exerts that force. B) What is the acceleration of the system? C) What is the tension at the top of the heavy rope? D) What is the tension at the midpoint of the rope? 200 N 6 kg 4 kg 5 kg Free Body Diagrams 6 kg block 200 N Ta w6 kg Tb 4 kg rope 5 kg block Ta Tb w4 kg w5 kg Calculating acceleration 6+5+4 kg blocks 200 N m a net F m a net 200 w eight m 6 5 4 15 kg w6+5+4 kg 15 a net 200 15 * 9.8 a net 200 15 * 9.8 15 3.53 m / s 2 Solving for tension at the top of the rope 6 kg block 200 N Ta Tb w6 kg 4 kg rope 5 kg block Ta Tb w4 kg For the 6 kg block, the net force is 6*3.53 so ma F 6 * 3.53 200 T a 6 * 9.8 T a 120 N w5 kg Solving for tension at the middle of the rope 6 kg block +2 kg (1/2 of rope) 200 N Ta w6 kg For the 6+2 kg, the net force is 8*3.53 so ma F 8 * 3 .5 3 2 0 0 T a 8 * 9 .8 T a 9 3 .3 N