j
Fy (T1 sin 30- ) 125 lbf 0
0.5 T1 125 lbf
T1 250 lbf
j
Forces
F1
F2
N
W
0
Application of Newton's Laws
FORCE EQUILIBRIUM
Rev. 0
Page 15
CP-04
Figure 8 Free-Body Diagram
By choosing (+) as the upward direction and (-) as the downward direction, the student can
determine by examination that 1) the upward component of T is + T sin 30 , 2) the tension T
1
1
3
o
is -125 lbf, and 3) T has no y- component. Therefore, using the same equation as before, we
2
obtain the following.
If the sum of all forces acting upon a body is equal to zero, that body is said to be in force
equilibrium. If the sum of all the forces is not equal to zero, any force or system of forces capable
of balancing the system is defined as an equilibrant.
Example:
A 2000 lbm car is accelerating (on a frictionless surface) at a rate of 2 ft-sec. What force
must be applied to the car to act as an equilibrant for this system?
Solution:
a.
Draw a free-body diagram.
b.
A Force, F , MUST be applied in the opposite direction to F such that the
2
1
sum of all forces acting on the car is zero.
