Question

*4 - 4. determine the moment of each of the three forces about point a.
4 - 5. determine the moment of each of the three forces about point b.
problems 4 - 4/5
f1 = 250 n 30°
f2 = 300 n
60°
2 m 3 m
4 m
f3 = 500 n

Explanation:

Step1: Recall moment formula

The moment of a force $F$ about a point is $M = r\times F$, where $r$ is the position - vector from the point to the point of application of the force. In scalar form, $M = Fd$, where $d$ is the perpendicular distance from the point to the line of action of the force.

Step2: Moment of $F_1$ about point $A$

For $F_1 = 250\ N$, the perpendicular distance from point $A$ to the line of action of $F_1$ is $d_1=2\sin30^{\circ}\ m$.
$M_{A1}=F_1\times d_1 = 250\times2\sin30^{\circ}=250\times2\times0.5 = 250\ N\cdot m$ (clock - wise)

Step3: Moment of $F_2$ about point $A$

The horizontal distance from $A$ to the point of application of $F_2$ is $x = 2 + 3=5\ m$ and the vertical distance is $y = 4\ m$. The perpendicular distance $d_2$ from $A$ to the line of action of $F_2$ is $d_2=(2 + 3)\sin60^{\circ}+4\cos60^{\circ}$.
$M_{A2}=F_2\times d_2=300\times((2 + 3)\sin60^{\circ}+4\cos60^{\circ})=300\times(5\times\frac{\sqrt{3}}{2}+4\times\frac{1}{2})=300\times(\frac{5\sqrt{3}}{2}+2)\approx300\times(4.33 + 2)=1959\ N\cdot m$ (clock - wise)

Step4: Moment of $F_3$ about point $A$

The perpendicular distance from $A$ to the line of action of $F_3$ is $d_3=(2 + 3)+4\times\frac{3}{5}$ (using similar - triangles for the vertical and horizontal components of the distance).
$M_{A3}=F_3\times d_3=500\times((2 + 3)+4\times\frac{3}{5})=500\times(5 + 2.4)=3700\ N\cdot m$ (clock - wise)

Answer:

$M_{A1}=250\ N\cdot m$ (clock - wise), $M_{A2}\approx1959\ N\cdot m$ (clock - wise), $M_{A3}=3700\ N\cdot m$ (clock - wise)

For the moment about point $B$:

Step1: Moment of $F_1$ about point $B$

The perpendicular distance from point $B$ to the line of action of $F_1$ is $d_{B1}=(2 + 3)\sin30^{\circ}+4\cos30^{\circ}$.
$M_{B1}=F_1\times d_{B1}=250\times((2 + 3)\sin30^{\circ}+4\cos30^{\circ})=250\times(5\times\frac{1}{2}+4\times\frac{\sqrt{3}}{2})=250\times(2.5 + 3.464)\approx1491\ N\cdot m$ (counter - clockwise)

Step2: Moment of $F_2$ about point $B$

The perpendicular distance from point $B$ to the line of action of $F_2$ is $d_{B2}=4\sin60^{\circ}$
$M_{B2}=F_2\times d_{B2}=300\times4\sin60^{\circ}=300\times4\times\frac{\sqrt{3}}{2}=600\sqrt{3}\approx1039\ N\cdot m$ (counter - clockwise)

Step3: Moment of $F_3$ about point $B$

The perpendicular distance from point $B$ to the line of action of $F_3$ is $0$ (since the line of action of $F_3$ passes through $B$).
$M_{B3}=0\ N\cdot m$