Question

A block of mass 8 kg is suspended by a rope of length 3 m from the ceiling. A horizontal force of 40 N is applied to the block. Determine the angle that the rope makes with the vertical in equilibrium. (Acceleration due to gravity \(g = 10 \, \text{m/s}^2\), neglect the mass of the rope)

• A. \(\sin^{-1} \frac{1}{2}\)
• B. \(\tan^{-1} \frac{1}{2}\)
• C. \(\sin^{-1} \frac{1}{3}\)
• D. \(\tan^{-1} \frac{1}{3}\)

Hint:

For equilibrium problems with a rope and horizontal force, resolve tension into vertical and horizontal components; the angle is determined using \(\tan \theta = F_{\text{horizontal}} / F_{\text{vertical}}\).

Answer 1

The Correct Option is B

Step 1: Picture the three forces as a closed triangle.
At equilibrium the block feels its weight $mg$ pulling straight down, the horizontal push $F$, and the rope tension $T$ along the rope. Since these three must sum to zero, they form a closed triangle when drawn tip to tail, and because weight is vertical while the push is horizontal, that triangle is a right triangle.

Step 2: Identify the sides of the triangle.
The vertical side has length $mg$, the horizontal side has length $F$, and the tension $T$ is the hypotenuse, making angle $\theta$ with the vertical side, which is the rope's angle from the vertical.

Step 3: Read the angle straight off the triangle.
In a right triangle, the tangent of the angle between the hypotenuse and a side equals the opposite side over the adjacent side:
\[ \tan\theta = \frac{F}{mg} = \frac{40}{8\times10} = \frac{40}{80} = \frac{1}{2} \]

Step 4: Solve for $\theta$.
\[ \theta = \tan^{-1}\frac{1}{2} \]

Step 5: Conclusion.
\[ \boxed{\tan^{-1}\frac{1}{2}} \]