Question

For the following question, enter the correct numerical value upto TWO decimal places. If the numerical value has more than two decimal places, round-off the value to TWO decimal places. (For example: Numeric value 5 will be written as 5.00 and 2.346 will be written as 2.35) A mass of 10 kg is suspended vertically by a rope from the roof. When a horizontal force is applied on the rope at some point, the rope deviates at an angle of \(45^\circ\) at that point. If the suspended mass is at equilibrium, the magnitude of the force applied is ____ N. (\(g = 10 \, \text{m s}^{-2}\))

Hint:

In equilibrium problems involving ropes, resolve tensions into horizontal and vertical components and apply the condition that net force in each direction is zero.

Answer 1

Correct Answer: 100

Step 1: Weight of the suspended mass: \[ W = mg = 10 \times 10 = 100 \text{ N} \] This is the tension in the vertical part of the rope.
Step 2: Let the tension in the inclined part of the rope be \(T\). Since the rope makes an angle of \(45^\circ\) with the vertical and the system is in equilibrium, the vertical component of \(T\) balances the weight: \[ T \cos 45^\circ = 100 \]
Step 3: \[ T = \frac{100}{\cos 45^\circ} = \frac{100}{\frac{1}{\sqrt{2}}} = 100\sqrt{2} \]
Step 4: The horizontal force applied equals the horizontal component of the tension: \[ F = T \sin 45^\circ \]
Step 5: \[ F = 100\sqrt{2} \times \frac{1}{\sqrt{2}} = 100 \text{ N} \] Final Answer (up to two decimal places): \[ \boxed{100.00} \]