A body is pulled on a horizontal surface with a rope inclined at 30º to the vertical. If the effective force…

Analysis

To solve this problem, we must resolve the tension in the rope into its horizontal and vertical components. A common mistake is to assume the angle given is always with the horizontal, but here, the angle is specified with the vertical.

Concept

When a force (Tension, $T$) is applied at an angle $\theta$ to the vertical:

• The vertical component is $T \cos \theta$.
• The horizontal component is $T \sin \theta$.

We are given:

• Angle with the vertical ($\theta$) = $30^\circ$
• Effective horizontal force ($F_h$) = $15\text{ N}$

Solution

1. Identify the formula: Since the horizontal component is $F_h = T \sin \theta$: $$15 = T \sin(30^\circ)$$

2. Substitute the value of $\sin(30^\circ)$: We know that $\sin(30^\circ) = 0.5$ (or $\frac{1}{2}$). $$15 = T \times 0.5$$

3. Solve for $T$: $$T = \frac{15}{0.5}$$ $$T = 30\text{ N}$$

The tension in the rope is $30\text{ N}$.

Conclusion

Comparing our result with the options provided, the correct answer is A.

Exam Tip

Always sketch a quick diagram! If the question says "inclined at $\theta$ to the vertical," draw a vertical line and the rope, then label the angle between them. Remember: the side opposite the angle always uses sine, and the side adjacent to the angle always uses cosine. Because the horizontal force is opposite the vertical angle, we correctly used sine.