Question

1. a 150 kg object is suspended from a ceiling and attached to a wall. what is the tension in the left - hand rope? a. 7.4×10² n b. 8.5×10² n c. 1.3×10³ n d. 2.5×10³ n

Explanation:

Step1: Calculate the weight of the object

The weight of the object $W = mg$, where $m = 150\ kg$ and $g=9.8\ m/s^{2}$. So $W=150\times9.8 = 1470\ N$.

Step2: Analyze the vertical - direction equilibrium

Let the tension in the left - hand rope be $T_1$ and the tension in the slanted rope be $T_2$. In the vertical direction, the vertical component of $T_2$ balances the weight of the object. $T_2\sin60^{\circ}=W$. So $T_2=\frac{W}{\sin60^{\circ}}=\frac{1470}{\sin60^{\circ}}$.

Step3: Analyze the horizontal - direction equilibrium

In the horizontal direction, $T_1 = T_2\cos60^{\circ}$. Substitute $T_2=\frac{W}{\sin60^{\circ}}$ into it, we get $T_1=\frac{W\cos60^{\circ}}{\sin60^{\circ}}=\frac{W}{\tan60^{\circ}}$.
Substitute $W = 1470\ N$ into the formula, $T_1=\frac{1470}{\tan60^{\circ}}\approx850\ N = 8.5\times 10^{2}\ N$.

Answer:

B. $8.5\times 10^{2}\ N$