Question

In Fig 12-66, asphere is supported on a frictionless plane inclined at angle $\theta =45°$from the horizontal. Angle $\varphi$is $25°$. Calculate the tension in the cable.

Short answer

Tension in the cable is$76N.$

Step-by-step solution

Listing the given quantities

$\text{θ}=45°$

$\varphi =25°$

Mass of sphere$\text{m}=10kg$

Understanding the concept of force and tension

We draw the free body diagram for the sphere. From this, we take the summation of all forces in the direction parallel to the ramp. After solving the equation, we will get the tension in the cable.

Equation:

${\sum }_{}^{}{\text{F}}_{\text{x}}=0$

Free Body Diagram

Calculation of tension in the cable

$\begin{array}{c}{\sum }_{}^{}{\text{F}}_{\text{alongtheinclinedplane}}=0\\ (\text{Tcos}\varphi )-(\text{mgsin}\theta )=0\\ (\text{Tcos}\varphi )=(\text{mgsin}\theta )\\ T=\frac{(\text{mgsin}\theta )}{\cos \varphi }\\ T=\frac{10kg\times 9.8m/s^2\times \sin 45°}{\cos 25°}\\ T=76N\end{array}$

Tension in the cable is$76N$