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Question: A roller of mass 300 kg and of radius 50 cm lying on horizontal floor is resting against a step of h...

A roller of mass 300 kg and of radius 50 cm lying on horizontal floor is resting against a step of height 20 cm. The minimum horizontal force to be applied on the roller passing through its centre to turn the roller on to the step is :

A

980 N

B

1960N

C

2940 N

D

3920 N

Answer

3920 N

Explanation

Solution

To find the minimum horizontal force required to turn the roller onto the step, we consider the torques about the pivot point. The roller is just about to turn, meaning it will pivot around the top edge of the step.

  1. Identify the pivot point: The pivot point (P) is the top edge of the step.

  2. Identify the forces causing torque about P:

    • Applied horizontal force (F): This force is applied at the center of the roller (C) and acts horizontally.
    • Weight of the roller (mg): This force acts vertically downwards through the center of the roller (C).

  3. Determine the lever arms for each force with respect to the pivot point P:

    • Let R be the radius of the roller, and h be the height of the step. Given: R = 50 cm = 0.5 m Given: h = 20 cm = 0.2 m Given: mass (m) = 300 kg Acceleration due to gravity (g) = 9.8 m/s²

    • Lever arm for the applied force (F): The force F is horizontal and applied at the center C. The perpendicular distance from the pivot point P to the line of action of F is the vertical distance between the center of the roller (C) and the pivot point (P). Vertical distance = R - h = 50 cm - 20 cm = 30 cm = 0.3 m. So, Torque due to F (τ_F) = F × (R - h)

    • Lever arm for the weight (mg): The weight mg acts vertically downwards through the center C. The perpendicular distance from the pivot point P to the line of action of mg is the horizontal distance from the pivot point (P) to the vertical line passing through the center of the roller (C). Let this horizontal distance be 'x'. We can find 'x' using the Pythagorean theorem by considering the right-angled triangle formed by the center of the roller (C), the pivot point (P), and the point directly below C on the horizontal level of P. The hypotenuse of this triangle is the radius R (distance from C to P). One leg is the vertical distance (R - h). The other leg is the horizontal distance 'x'. So, x² + (R - h)² = R² x² + (50 cm - 20 cm)² = (50 cm)² x² + (30 cm)² = (50 cm)² x² + 900 = 2500 x² = 1600 x = 40 cm = 0.4 m. So, Torque due to mg (τ_mg) = mg × x

  4. Apply the condition for minimum force: For the roller to just turn onto the step, the torque due to the applied force must be equal to the torque due to its weight. τ_F = τ_mg F × (R - h) = mg × x F × 0.3 m = (300 kg) × (9.8 m/s²) × 0.4 m

  5. Solve for F: F = (300 × 9.8 × 0.4) / 0.3 F = (300 × 9.8 × 4) / 3 F = 100 × 9.8 × 4 F = 980 × 4 F = 3920 N

The minimum horizontal force required is 3920 N.