Question

consider the figure below. (a) find the tension in each cable supporting the 700 n cat burglar. (assume the angle θ of the inclined cable is 36.0°.) inclined cable horizontal cable vertical cable (b) suppose the horizontal cable were re - attached higher up on the wall. would the tension in the inclined cable increased, decrease, or stay the same? increase decrease stay the same

Explanation:

Step1: Analyze vertical - force equilibrium for part (a)

Let the tension in the inclined cable be $T$. In the vertical - direction, the vertical component of the tension in the inclined cable supports the weight of the cat burglar. So, $T\sin\theta = W$, where $W = 700\ N$ and $\theta=36.0^{\circ}$. Then $T=\frac{W}{\sin\theta}$.

Step2: Calculate the tension in the inclined cable

$T=\frac{700}{\sin(36.0^{\circ})}=\frac{700}{0.5878}\approx1191\ N$. In the horizontal - direction, the tension in the horizontal cable $T_h$ is equal to the horizontal component of the tension in the inclined cable. So, $T_h = T\cos\theta$. Substituting $T=\frac{700}{\sin\theta}$, we get $T_h=\frac{700\cos\theta}{\sin\theta}=700\cot\theta$. Then $T_h = 700\cot(36.0^{\circ})=700\times1.3764\approx963\ N$. The tension in the vertical cable is equal to the weight of the cat burglar, so $T_v = 700\ N$.

Step3: Analyze the effect of re - attaching the horizontal cable for part (b)

The weight of the cat burglar is still supported by the vertical component of the tension in the inclined cable, $T\sin\theta = W$. If the horizontal cable is re - attached higher up on the wall, the angle $\theta$ of the inclined cable with the horizontal increases. Since $T=\frac{W}{\sin\theta}$, and as $\theta$ increases from the original value, $\sin\theta$ increases (for $\theta$ in the range $0^{\circ}<\theta < 90^{\circ}$). So, the tension in the inclined cable $T$ decreases.

Answer:

(a) Tension in inclined cable: $1191\ N$, Tension in horizontal cable: $963\ N$, Tension in vertical cable: $700\ N$
(b) Decrease