Question

3.4 a 300 - n force is applied at a as shown. determine (a) the moment of the 300 - n force about d, (b) the smallest force applied at b that creates the same moment about d. fig. p3.4 and p3.5

Explanation:

Step1: Resolve the 300 - N force into components

The 300 - N force is at an angle of 25° with the vertical. The horizontal component $F_x=300\sin25^{\circ}$ and the vertical component $F_y = 300\cos25^{\circ}$.

Step2: Calculate the moment of the 300 - N force about point D

The perpendicular distance from point D to the line of action of the horizontal component $d_x=200$ mm and to the vertical component $d_y = 100$ mm.
The moment of a force $M$ about a point is given by $M=\sum(F\times d)$.
$M = F_x\times100+F_y\times200$.
$F_x = 300\sin25^{\circ}\approx300\times0.4226 = 126.78$ N.
$F_y=300\cos25^{\circ}\approx300\times0.9063 = 271.89$ N.
$M=(126.78\times100 + 271.89\times200)\times10^{-3}$ N·m.
$M=(12678+54378)\times10^{-3}$ N·m.
$M = 67.056$ N·m.

Step3: Find the smallest force at B to create the same moment

The smallest force at B will act perpendicular to the line joining B and D. The distance from B to D, using the Pythagorean theorem, $r=\sqrt{(200 - 125)^2+(200 + 100)^2}=\sqrt{75^2+300^2}=\sqrt{5625 + 90000}=\sqrt{95625}=309.23$ mm.
Since $M = F\times d$, and we want $M = 67.056$ N·m and $d$ is the perpendicular distance from B to D. The smallest force $F=\frac{M}{d}$.
$F=\frac{67.056}{0.30923}\approx216.85$ N.

Answer:

(a) The moment of the 300 - N force about D is $67.06$ N·m (rounded to two decimal places).
(b) The smallest force applied at B that creates the same moment about D is approximately $216.9$ N (rounded to one decimal place).