A block weighing 400 kg rest on a horizontal surface and supports on top of it another block of weight 100 kg placed on the top of it as shown. The block W2 is attached to a vertical wall by a string 6 m long. Ifthe coefficient of friction between all surfaces is 0.25 and the system is in equilibrium find the magnitude of the horizontal force P applied to the lower block.

The horizontal force applied to the lower block is approximately 1,420.85 Newtons

The known parameters are;

The mass of the block, m₁ = 400 kg, weight, W₁ = 3,924 N

The mass of the block resting on the first block, m₂ = 100 kg, weight, W₂ = 981 N

The length of the string attached to the block, W₂, l = 6 m

The horizontal distance from the point of attachment of the second block to the block W₂, x = 5 m

The coefficient of friction between the surfaces, μ = 0.25

Let T represent the tension in the string

The upward force on W₂ due to the string = T × sin(θ)

The normal force of W₁ on W₂, N₂ = W₂ - T × sin(θ)

The tension in the string, T = N₂ × μ × cos(θ)

∴ T = (W₂ - T × sin(θ)) × μ × cos(θ)

sin(θ) = √(6² - 5²)/6

cos(θ) = 5/6

∴ T = (981 - T × √(6² - 5²)/6) × 0.25 × 5/6

Solving, we get;

T ≈ 183.27 N

The normal reaction on W₂, N₂ = T/(μ × cos(θ))

∴ N₂ = 183.27/(0.25 × 5/6) = 879.7

N₂ ≈ 879.7 N

The friction force, = N₂ × μ

∴ = 879.7 N × 0.25 = 219.925 N

The total normal reaction on the ground, = W₁ + N₂

= 3,924 N + 879.7 N = 4,803.7 N

The friction force, on the ground = × μ

∴   = 4,803.7 N × 0.25 = 1,200.925 N

The horizontal force applied to the lower block, P = +

Therefore;

P = 1,200.925 N + 219.925 N = 1,420.85 N

The horizontal force applied to the lower block, P ≈ 1,420.85 N

i had this question and the answer was 3.15 miles.

martina travels from school to her mom's office. from there, she walks to school. after school, she walks to the library and then to a friends house before finally ending back at home. find martina's total distance traveled.

a) 0 miles  

b) 0.4 miles  

c) 1 mile  

d) 3.15 miles