Question

The horizontal position of the 500.kg rectangular block of concrete is adjusted by the 5° wedge under the action of the force p. if the coefficient of static friction for both pairs of wedge sur faces is 0.30 and if the coefficient of static friction between the block and the horizontal surface is 0.60, determine the least force p required to move the block. 2.​

Answer 1

Given:

1. Weight of concrete block = 500kg
2. Inclination of wedge = 5°
3. Coefficient of static friction of wedge surfaces = 0.30
4. Coefficient of static friction between the block and the horizontal surface = 0.60

To find:

The least force P required to move the block

Answer:

• Let ω denote the coefficient of friction between the block and the wedge and Ф denote the inclination of wedge
• ω = (Coefficient of friction between the block and the ground/ Coefficient of friction between the wedge surface)
• ω = 0.3/0.6 = 0.5

The forces acting on the block are -

• Weight W (↓) = mg = 500* 9.82 = 4910N
• Frictional force F ( along the slope) = ω*mg = 0.5 *4910 = 2455N
• Minimum force required to move the block P
• On resolving the forces along the axes of the block we get,

1. P + WsinФ = F
2. P = F - WsinФ
3. P = 2455 - 4910 sin(5°) = 2455 - 429.1 = 2025.9N

The least force P required to move the block is 2025.9N