Question

<hw 4: scalar moments adaptive follow - up
problem 4.86

two couples act on the frame shown in (figure 1).

part a
if d = 3.6 ft, determine the resultant couple moment either by summing the moments of the two couples or resolving each force into x and y components and then summing the moments about a of all the force components.
express your answer in pound - feet to three significant figures. assume the positive direction is counterclockwise.
mr = - 180 lb - ft
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figure
< 1 of 1 >

100 lb
30°
150 lb
3 ft
3 ft
b
3
4 ft
30°
100 lb
150 lb

Explanation:

Step1: Resolve forces into components

For the 100 - lb force:
The vertical component $F_{1y}=100\sin30^{\circ}= 50$ lb and the horizontal component $F_{1x}=100\cos30^{\circ}=50\sqrt{3}$ lb.
For the 150 - lb force:
The vertical component $F_{2y}=150\times\frac{3}{5}=90$ lb and the horizontal component $F_{2x}=150\times\frac{4}{5}=120$ lb.

Step2: Calculate moment due to 100 - lb couple

The moment of the 100 - lb couple about point $A$, $M_1 = 100\times(3 + d)\sin30^{\circ}$. Substituting $d = 3.6$ ft, we get $M_1=100\times(3 + 3.6)\times\frac{1}{2}=330$ lb - ft (counter - clockwise).

Step3: Calculate moment due to 150 - lb couple

The moment of the 150 - lb couple about point $A$, $M_2=-150\times3\times\frac{3}{5}=-270$ lb - ft (clockwise).

Step4: Sum the couple moments

$M_R = M_1+M_2$.
$M_R=330-270 = 60.0$ lb - ft.

Answer:

$60.0$ lb - ft