A concrete block of mass 35kg is pulled along a horizontal floor with the aid of a rope inclined at an angle of 30 degrees to the horizontal. If the coefficient of the friction is 0.75. Calculate the force required to move the block over the floor
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To calculate the force required to move the block over the floor, we need to consider the forces acting on the block. There are three main forces at play: the gravitational force (mg), the normal force (N), and the force of friction (Ff).
1. Gravitational force (mg): This force is the product of the mass of the block (m) and the acceleration due to gravity (g ≈ 9.8 m/s²). In this case, the gravitational force is given by Fg = mg.
Fg = 35 kg × 9.8 m/s² = 343 N
2. Normal force (N): It is the force exerted by a surface to support the weight of an object resting on it. For an object on a horizontal surface, the normal force is equal in magnitude but opposite in direction to the gravitational force. So in this case, the normal force is equal to Fg.
N = Fg = 343 N
3. Force of friction (Ff): The force of friction resists the motion of the block and is determined by the coefficient of friction (μ) and the normal force (N). The force of friction can be calculated using the equation Ff = μN.
Ff = 0.75 × 343 N = 257.25 N
Therefore, the force required to move the block over the floor is approximately 257.25 N.
M*g = 35*9.8 = 343 N. = Wt. of block. = Normal force, Fn.
Fs = u*Fn = 0.75*343 = 257.3 N. = force of starting friction.
F*Cos30 - Fs = M*a.
F*Cos30 - 257.3 = 35*0.
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