Chapter 6.3, Problem 6.87P

Determine the components of the reactions at A and B, (a) if the 60-lb load is applied as shown, (b) if the 60-lb load is moved along its line of action and is applied at point E.

Fig. P6.87

SOLUTION

Free body: Entire frame:

Analysis is valid for either parts (a) or (b), since position of 60-lb load on its line of action is immaterial.

∑M_B = 0: − A_y (5 in.) – (60 lb)(4 in.) = 0 A_y = −48 lb

+↑ ∑F_y = 0: B_y − 60 − 480 = 0 B_y = +108 lb

∑F_x = 0: A_x + B_x = 0    (1)

a) Load applied at D.

Since BCE is a two-force member (with forces applied at B and C only), the reaction at B must be directed along CB. We have, therefore,

B_x = 0

The reaction at B is B = 108.0 lb↑

From Eq. (1): A_x + 0 = 0 A_x = 0

The reaction at A is A = 48.0 lb↓

(b) Load applied at E.

Since ACD is a two-force member, the reaction at A must be directed along AC. We have

$\frac{3\text{ in}\text{.}}{\text{5 in}\text{.}}=\frac{48\text{ lb}}{A_x}$ A_x = 80.0 lb →, A_y = 48.0 lb ↓

From Eq. (1): 80 lb + B_x = 0 B_x = 80 lb

Thus, B_x = 80.0 lb ←, B_y = 108.0 lb ↑