ENGR-1100 Introduction to
Engineering Analysis
Lecture 24
Notes courtesy of: Prof. Yoav Peles
Lecture outline
• Machines
Class Example:
A pin-connected system of bars supports a 300 lb
load as shown. Determine the reactions at supports
A and B and the force exerted by the pin at C on
member ACE.
Solution
From a free-body diagram
For the complete system:
+ MA = - Bx (20) - 300 (30) = 0
Bx = -450 lb = 450 lb 
+  Fx = Ax + Bx = Ax - 450 = 0
Ax = 450 lb = 450 lb 
From a free-body diagram
for pin F (note use of local
coordinates):
+  Fy’ = TEF -300 sin 45 = 0
TEF =212.1 lb  212 lb (T)
From a free-body diagram
for bar ACE:
+ MC = -Ay (10) + 450 (10)
- 212.1 (10/cos 45) = 0
Ay = 150.0 lb = 150.0 lb 
+  Fy = Ay + Cy - 212.1 sin 45
= 150.0 + Cy - 212.1sin 45 = 0
Cy = 0
+  Fx = Cx+ 450+212.1 cos 45=0
Cx = - 600 lb
From a free-body diagram
For the complete system:
+  Fy = By + Ay - 300
= By + 150.0 - 300 = 0
By = 150.0 lb = 150.0 lb 
A=
A   A  =
A =
2
2
x
tan-1
y
 450.0   150.0  = 474.3 lb  474 lb
150.0
= 18.434
450.0
2
2
A  474 lb 18.43
Example
Forces of 50 lb are applied to the handles of the bolt
cutter. Determine the force exerted by on the bolt at E
and all forces acting on the handle ABC.
Solution
From a free-body diagram
for member CDE:
+  Fx = Cx = 0
Cx = 0
+ MD = Cy (3) - E(2) = 0
E
3
Cy
2
From a free-body diagram
For handle ABC:
+ MB = Cy (1) - 50 (20) = 0
Cy =1000 lb = 1000 lb 
Cy = 1000 lb 
Fx = Bx = 0
Bx = 0
Fy = By + 50 + 1000 = 0
By = -1050 lb
E = 1.5Cy = 1.5 (1000) = 1500 lb
Force on the bolt:
By = 1050 lb 
E = 1500 lb 