Question

in (figure 1), the cylinder d has a mass of 20 kg. part a if a force of f = 100 n is applied horizontally to the ring at a, determine dimension d so that the force in cable ac is zero. express your answer to three significant figures and include the appropriate units.

Explanation:

Step1: Calculate the weight of cylinder D

The weight of cylinder D is given by $W = mg$, where $m = 20\ kg$ and $g=9.81\ m/s^{2}$. So $W=20\times9.81 = 196.2\ N$.

Step2: Set up equilibrium equations

Since the force in cable AC is zero, consider the equilibrium of forces at point A in the horizontal and vertical - directions. Let the tension in cable AB be $T_{AB}$. In the horizontal direction, $\sum F_x=F - T_{AB}\cos\theta=0$, where $F = 100\ N$. In the vertical direction, $\sum F_y=T_{AB}\sin\theta - W = 0$. From $\sum F_y=T_{AB}\sin\theta - W = 0$, we get $T_{AB}=\frac{W}{\sin\theta}$. Substituting into $\sum F_x$: $F-\frac{W}{\sin\theta}\cos\theta = 0$, and since $\frac{\cos\theta}{\sin\theta}=\cot\theta$, we have $F - W\cot\theta=0$. Then $\cot\theta=\frac{F}{W}$.

Step3: Relate the angle to the dimension d

We know that $\cot\theta=\frac{d}{2}$. Substituting $F = 100\ N$ and $W = 196.2\ N$ into $\cot\theta=\frac{F}{W}$, we get $\cot\theta=\frac{100}{196.2}$. And since $\cot\theta=\frac{d}{2}$, then $d = 2\times\frac{100}{196.2}$.

Answer:

$d = 1.02\ m$