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 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)
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For equilibrium problems with a rope and horizontal force, resolve tension into vertical and horizontal components; the angle is determined using \(\tan \theta = F_{\text{horizontal}} / F_{\text{vertical}}\).
Updated On: Jul 18, 2026
- \(\sin^{-1} \frac{1}{2}\)
- \(\tan^{-1} \frac{1}{2}\)
- \(\sin^{-1} \frac{1}{3}\)
- \(\tan^{-1} \frac{1}{3}\)
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The Correct Option is B
Solution and Explanation
Step 1: Identify forces acting on the block.
The block of mass \(m = 8 \, \text{kg}\) experiences gravity \(mg\) downward, tension \(T\) along the rope, and a horizontal applied force \(F = 40 \, \text{N}\).
Step 2: Resolve tension into components.
Let the rope make an angle \(\theta\) with the vertical. Then the tension \(T\) has:
- Vertical component: \(T \cos \theta = mg\) (balances weight)
- Horizontal component: \(T \sin \theta = F\) (balances applied horizontal force)
Step 3: Express angle using tangent.
\[ \tan \theta = \frac{\text{horizontal component}}{\text{vertical component}} = \frac{F}{mg} = \frac{40}{8 \times 10} = \frac{40}{80} = \frac{1}{2} \]
Step 4: Solve for \(\theta\).
\[ \theta = \tan^{-1} \frac{1}{2} \]
Step 5: Verification.
Check: vertical component \(T \cos \theta = mg = 80 \, \text{N}\), horizontal \(T \sin \theta = F = 40 \, \text{N}\). Ratio \(\tan \theta = 40/80 = 1/2\), consistent.
Step 6: Final conclusion.
Hence, the angle the rope makes with the vertical is:
\[ \boxed{\tan^{-1} \frac{1}{2}} \]
The block of mass \(m = 8 \, \text{kg}\) experiences gravity \(mg\) downward, tension \(T\) along the rope, and a horizontal applied force \(F = 40 \, \text{N}\).
Step 2: Resolve tension into components.
Let the rope make an angle \(\theta\) with the vertical. Then the tension \(T\) has:
- Vertical component: \(T \cos \theta = mg\) (balances weight)
- Horizontal component: \(T \sin \theta = F\) (balances applied horizontal force)
Step 3: Express angle using tangent.
\[ \tan \theta = \frac{\text{horizontal component}}{\text{vertical component}} = \frac{F}{mg} = \frac{40}{8 \times 10} = \frac{40}{80} = \frac{1}{2} \]
Step 4: Solve for \(\theta\).
\[ \theta = \tan^{-1} \frac{1}{2} \]
Step 5: Verification.
Check: vertical component \(T \cos \theta = mg = 80 \, \text{N}\), horizontal \(T \sin \theta = F = 40 \, \text{N}\). Ratio \(\tan \theta = 40/80 = 1/2\), consistent.
Step 6: Final conclusion.
Hence, the angle the rope makes with the vertical is:
\[ \boxed{\tan^{-1} \frac{1}{2}} \]
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