Examples from Chapter 4

1. Examples from Chapter 4

2. 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.

3. R F2 q F1 Step 1 Draw It!

4. R F2 q F1 Step 2 Break Forces into Components F2 sin (1800-q) F2 sin (1800-q)=F2 sin (q) F2 cos (1800-q)=-F2 cos (q)

5. 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

6. Problem 4.49 200 N 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. • 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. • What is the acceleration of the system? • What is the tension at the top of the heavy rope? • What is the tension at the midpoint of the rope? 6 kg 4 kg 5 kg

7. 6 kg block 4 kg rope 5 kg block 200 N Ta Tb Ta Tb w6 kg w4 kg w5 kg Free Body Diagrams

8. Calculating acceleration 6+5+4 kg blocks 200 N w6+5+4 kg

9. 6 kg block 4 kg rope 5 kg block 200 N Ta Tb Ta Tb w6 kg w4 kg w5 kg Solving for tension at the top of the rope For the 6 kg block, the net force is 6*3.53 so

10. 6 kg block +2 kg (1/2 of rope) 200 N Ta w6 kg Solving for tension at the middle of the rope For the 6+2 kg, the net force is 8*3.53 so