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. Step 1 Draw It! F1 R F2 q

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

5. In the horizontal direction: F1-F2cos(q)=0 In the vertical direction:
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
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?
200 N
6 kg
4 kg
5 kg

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

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

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

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