Question

A garden roller of weight \( 100\,kg \) is pulled with a force of \( 300\,N \) acting at an angle of \( 30^\circ \) with the ground. The effective pulling weight of the roller (in kg wt) is \((g=10\,m/s^2)\)

• A. \( 850 \)
• B. \( 725 \)
• C. \( 800 \)
• D. \( 820 \)
• E. \( 700 \)

Hint:

When a force is applied at an angle, its vertical component reduces or increases the effective weight depending on its direction.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The "effective pulling weight" seems to refer to the normal force exerted by the roller on the ground. When the roller is pulled by a force at an angle, the vertical component of this force acts upwards, partially counteracting the roller's weight. This reduces the normal force. However, the term "effective pulling weight" is non-standard. A more likely interpretation, given the options, is that the question is asking for the effective weight of the roller as felt by the ground (i.e., the normal reaction force), but the values in the options are inconsistent with this interpretation (they are less than the actual weight). Let's re-read carefully: "effective pulling weight". This is highly ambiguous. Let's assume there is a typo and it means "effective vertical force" on the ground, or the normal force.
Re-interpretation based on ambiguity and options: The term "effective pulling weight" is extremely unusual. Let's consider what might be intended. It is possible the question is asking for the net downward force. Or, perhaps "pulling weight" is a convoluted way of asking for the normal force. Let's calculate the normal force. Step 2: Key Formula or Approach:
1. Draw a free-body diagram of the roller.
2. Resolve the pulling force into its horizontal and vertical components.
3. Apply the condition of vertical equilibrium (sum of vertical forces is zero) to find the normal force.
4. The weight of the roller is \( W = mg \).
5. The pulling force is \( F = 300 \) N at an angle \( \theta = 30^\circ \).
The vertical component of the pulling force is \( F_y = F \sin\theta \).
The horizontal component is \( F_x = F \cos\theta \).
Step 3: Detailed Explanation:
First, calculate the weight of the roller in Newtons. The mass is given as 100 kg (as weight is measured in N, "weight 100 kg" is a common colloquialism for "mass 100 kg").
\[ W = mg = 100 \text{ kg} \times 10 \text{ ms}^{-2} = 1000 \text{ N} \] The pulling force is \( F = 300 \) N at an angle of \( 30^\circ \) with the ground.
The vertical component of this force, \( F_y \), acts upwards:
\[ F_y = F \sin(30^\circ) = 300 \times \frac{1}{2} = 150 \text{ N} \] The forces acting in the vertical direction are:
- Weight (W) acting downwards.
- Vertical component of pulling force (\(F_y\)) acting upwards.
- Normal reaction force (N) from the ground acting upwards.
For vertical equilibrium, the net vertical force is zero:
\[ N + F_y - W = 0 \] \[ N = W - F_y \] \[ N = 1000 \text{ N} - 150 \text{ N} = 850 \text{ N} \] The normal force, or the effective force the roller exerts on the ground, is 850 N.
The question asks for the answer in "kg wt". 1 kg wt is the force due to gravity on a 1 kg mass, which is \( 1 \times g \). With g = 10, 1 kg wt = 10 N.
So, to convert our answer from Newtons to kg wt, we divide by g:
\[ \text{Normal force in kg wt} = \frac{850 \text{ N}}{10 \text{ N/kg wt}} = 85 \text{ kg wt} \] Looking at the options (850, 725, 800, 820, 700), none of them match 85. However, option (A) is 850, which is our answer in Newtons. It is highly probable that the question intended to ask for the effective weight (normal force) in Newtons but mistakenly wrote "kg wt". Assuming this typo, the answer is 850.
Step 4: Final Answer:
Assuming the question asks for the normal force in Newtons, the answer is 850 N. This corresponds to option (A).