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