9. A block with a mass of 2.0 kg is pulled across a horizontal plane by a horizontal force of 10 N. This force is just sufficient to overcome friction. Calculate the coefficient of friction between the block and the plane. 10. A block with mass (m) of 70 kg is pulled across a horizontal plane with a force (F) of 300 N. This force is just sufficient to overcome friction. Calculate the coefficient of friction between the block and the plane. 11. A block with a mass of 15 kg is pulled across a horizontal plane with a force of 44 N. Calculate the: (a) weight of the block (w). (b) coefficient of friction (μ) between the block and the horizontal plane. (c) pulling force required to pull a similar block of 2 kg mass under the same conditions. A force of 150 N is required to move a casting with a mass of 43 kg across a horizontal plane. Calculate the value of the coefficient of friction and the magnitude of the angle of rest. The coefficient of friction between a block and a horizontal plane is 0. Calculate the angle of rest.
Answer (1)
To solve these questions, we need to understand the concept of friction and specifically the coefficient of friction. The coefficient of friction (denoted as μ ) is a ratio that describes the force of friction between two bodies in contact relative to the normal force pressing them together.
9. Calculation for a 2 kg Block
Given:
Mass m = 2.0 kg
Force F = 10 N
Calculate the weight of the block: w = m ⋅ g = 2.0 ⋅ 9.8 = 19.6 N Where g is the acceleration due to gravity ( 9.8 m/s 2 ) .
The force of friction f is equal to the applied force since it's just sufficient to keep the block moving: f = F = 10 N
The frictional force f is also defined by μ ⋅ N , where N is the normal force, which is the same as the weight here: 10 = μ ⋅ 19.6
Solve for μ : μ = 19.6 10 = 0.51
10. Calculation for a 70 kg Block
Given:
Mass m = 70 kg
Force F = 300 N
Calculate the weight: w = m ⋅ g = 70 ⋅ 9.8 = 686 N
The force of friction f = 300 N .
Solve for μ : μ = 686 300 ≈ 0.44
11. Calculation for a 15 kg Block
Given:
Mass m = 15 kg
Force F = 44 N
(a) Weight w : w = 15 ⋅ 9.8 = 147 N
(b) Coefficient of friction μ : μ = 147 44 ≈ 0.30
(c) Pulling force for a 2 kg block:
Weight of the 2 kg block = 2 ⋅ 9.8 = 19.6 N
Required force F = μ ⋅ 19.6 = 0.30 ⋅ 19.6 = 5.88 N
The calculations follow the same principle where the frictional force must be equal to the applied force to maintain movement. These steps demonstrate the process of understanding how the coefficient of friction works between different blocks and applying forces across surfaces.
