1. Problem 4-d Rod AC is supported by a pin and
bracket at A and rests against a
peg at B. Neglecting the effect of
friction, determine (a) the reactions
at A and B when a = 8 in., (b) the
distance a for which the reaction
at A is horizontal and the
corresponding magnitudes of the
reactions at A and B. A B C a
20 in
10 in
60 lb

2. Problem 4-d A B C a
20 in
10 in
60 lb
Solving Problems on Your Own
Rod AC is supported by a pin and
bracket at A and rests against a
peg at B. Neglecting the effect of
friction, determine (a) the reactions
at A and B when a = 8 in., (b) the
distance a for which the reaction
at A is horizontal and the
corresponding magnitudes of the
reactions at A and B.
1. Draw a free-body diagram of the body. This diagram shows
the body and all the forces acting on it.

3. Problem 4-d A B C a
20 in
10 in
60 lb
Solving Problems on Your Own
Rod AC is supported by a pin and
bracket at A and rests against a
peg at B. Neglecting the effect of
friction, determine (a) the reactions
at A and B when a = 8 in., (b) the
distance a for which the reaction
at A is horizontal and the
corresponding magnitudes of the
reactions at A and B.
2. For a three-force body, solution can be obtained by
constructing a force triangle. The resultants of the three forces
must be concurrent or parallel. To solve a problem involving a
three-force body with concurrent forces, draw the free-body
diagram showing that the three forces pass through the same
point. Complete the solution by using a force triangle.

4. tan a = a = 26.57o Problem 4-d Solution A (a) a = 8 in aC a
20 in
10 in
60 lb
(a) a = 8 in
Draw a free-body diagram
of the body. C
10 in
60 lb A g a
8 in
12 in B 2 1 G D F E
tan a =
a = 26.57o 1 2

5. AE = EB = (8) = 4 in. EF = BG = 10 _ 4 = 6 in DG = BG = (6) = 3 in.C
10 in
60 lb A g a
8 in
12 in B 2 1 G D F E
Problem 4-d Solution
Construct a force triangle.
3 - FORCE BODY
Reaction at A passes through D
where B and 60-lb load intersect
AE = EB = (8) = 4 in.
EF = BG = 10 _ 4 = 6 in
DG = BG = (6) = 3 in.
FD = FG _ DG = 8 _ 3 = 5 in.
Tan g = = ; g = 26.57o 1 2 FD AF 5 10

6. 10 in
Problem 4-d Solution A
FORCE TRIANGLE g E F A
a = 26.57o
8 in D
30 lb A G
60 lb a B B
12 in B 2
30 lb
a = 26.57o 1 C
60 lb
10 in
A = B = = lb
A = 67.1 lb o
B = 67.1 lb o
sin 26.57o
30 lb

7. a Problem 4-d Solution A (b) For A horizontal aC a
20 in
10 in
60 lb
(b) For A horizontal
Draw a free-body diagram
of the body.
10 in A F a A a a G a B B 2
a = 26.57o 1 C
60 lb

8. a D ABF : BF = AF cos a D BFG : FG = BF sin a a = FG = AF cos a sin a
Problem 4-d Solution A
Construct a force triangle. F a A a a G a B
D ABF : BF = AF cos a
D BFG : FG = BF sin a
a = FG = AF cos a sin a
a = (10 in.) cos 26.57o sin 26.57o
a = 4.00 in. B 2
a = 26.57o 1 C
60 lb

9. 10 in
Problem 4-d Solution A
FORCE TRIANGLE F a A a a G a B
60 lb A B
a = 26.57o B 2
a = 26.57o 1 C
60 lb
tan a
60 lb
A = = 120 lb A = lb
B = = lb B = lb o
sin a
60 lb