Question

3.85 a worker tries to move a rock by applying a 360 - n force to a steel bar as shown. (a) replace that force with an equivalent force - couple system at d. (b) two workers attempt to move the same rock by applying a vertical force at a and another force at d. determine these two forces if they are to be equivalent to the single force of part a.

Explanation:

Step1: Resolve the given 360 - N force into components

The 360 - N force has a horizontal component $F_x = 360\sin40^{\circ}$ and a vertical component $F_y=360\cos40^{\circ}$.
$F_x = 360\sin40^{\circ}\approx360\times0.6428 = 231.4\ N$
$F_y = 360\cos40^{\circ}\approx360\times0.7660 = 275.8\ N$

Step2: Set up equations for equivalent force - couple system

Let the vertical force at $A$ be $F_A$ and the force at $D$ be $F_D$.
For the vertical equilibrium of the force - couple system, the sum of vertical forces must be equal to the vertical component of the 360 - N force. Let's assume upward as positive.
$\sum F_y=F_A + F_D\sin30^{\circ}=275.8$
For the moment equilibrium about a point (say $D$), the moment of the 360 - N force about $D$ must be equal to the moment of $F_A$ about $D$.
The moment of the 360 - N force about $D$:
$M_1 = 360\cos40^{\circ}(0.3 + 0.35)+360\sin40^{\circ}(0.4)$
$M_1=275.8\times0.65 + 231.4\times0.4$
$M_1 = 179.27+92.56=271.83\ N\cdot m$
The moment of $F_A$ about $D$ is $M_2 = F_A(0.3 + 0.35 + 0.4)$
$M_2 = F_A\times1.05$
Since $M_1 = M_2$, we have $F_A=\frac{271.83}{1.05}\approx258.9\ N$
Substitute $F_A$ into $\sum F_y$:
$258.9+F_D\sin30^{\circ}=275.8$
$F_D\sin30^{\circ}=275.8 - 258.9 = 16.9$
$F_D=\frac{16.9}{\sin30^{\circ}}=33.8\ N$

Answer:

The vertical force at $A$ is approximately $259\ N$ and the force at $D$ is approximately $33.8\ N$