Question

1. an object is suspended on a frictionless inclined place by a rope parallel to the incline as shown. if the angle of the incline is 25° and the tension in the rope is 5000 n, what is the weight of the object?

Explanation:

Step1: Resolve weight into components

Let the weight of the object be $W$. On a friction - less inclined plane, the tension $T$ in the rope parallel to the incline balances the component of the weight along the incline. The component of the weight along the incline is $W\sin\theta$, where $\theta = 25^{\circ}$ and $T$ is the tension in the rope.

Step2: Set up the equilibrium equation

Since the object is in equilibrium, $T = W\sin\theta$. We need to solve for $W$. Rearranging the equation gives $W=\frac{T}{\sin\theta}$.

Step3: Substitute the values

We know that $T = 5000\ N$ and $\theta=25^{\circ}$, and $\sin25^{\circ}\approx0.4226$. So $W=\frac{5000}{0.4226}\approx11831.52\ N$.

Answer:

$11831.52\ N$