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!

F1

RF2

Step 2 Break Forces into Components

F1

R

F2

F2 sin (1800-)

F2 cos (1800-)=-F2 cos ()F2 sin (1800-)=F2 sin ()

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()=0

In the vertical direction:

F2sin()=R=1300 N

So F2cos()=F1=1300 N

Thus, =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?

6 kg

5 kg

4 kg

200 N

Free Body Diagrams

w6 kg

Ta

200 N

6 kg block

w4 kg

Tb

Ta

4 kg rope

w5 kg

Tb

5 kg block

Calculating acceleration

w6+5+4 kg

200 N

6+5+4 kg blocks

2

200

6 5 4 15

15 200 15*9.8

200 15*9.83.53 /

15

net

net

net

net

ma F

ma weight

m kg

a

a m s

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

6*3.53 200 6*9.8

120a

a

ma F

T

T N

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

8*3.53 200 8*9.8

93.3a

a

ma F

T

T N