A pair of blocks of the mass M =30.7 kg, and 2 M are dragged over a rough, horizontal surface by two constant forces of magnitude F=98.0 N. The angle α = 22.7 ° . The blocks are displaced 7.5 m, and the coefficient of kinetic friction is 0.258.
*for the normal force of the back block I calculated 307N
*for the magnitude of friction experienced on the back block I calculated 79.206N
I need help calculating:
-the magnitude of the normal force exerted by the floor on the front block?
-the magnitude of the friction force experienced by the front block?
Please help! my homework is due at midnight and I can't figure this out.
To calculate the magnitude of the normal force exerted by the floor on the front block, we can use Newton's second law in the vertical direction. At equilibrium, the sum of the vertical forces is equal to zero.
Since the front block is not accelerating vertically, the normal force and weight must cancel each other out. We can express this as:
Normal force - Weight (front block) = 0
The weight of an object is given by the formula:
Weight = mass * gravitational acceleration
Given that the mass of the front block is M = 30.7 kg, and the acceleration due to gravity is approximately 9.8 m/s^2, we can substitute these values:
Normal force - (30.7 kg * 9.8 m/s^2) = 0
Simplifying, we find:
Normal force = (30.7 kg * 9.8 m/s^2)
Therefore, the magnitude of the normal force exerted by the floor on the front block is approximately 300.86 N.
To calculate the magnitude of the friction force experienced by the front block, we can use the formula:
Friction force = (coefficient of kinetic friction) * (normal force)
Given that the coefficient of kinetic friction is 0.258 and the normal force on the front block is approximately 300.86 N, we can substitute these values:
Friction force = (0.258) * (300.86 N)
Therefore, the magnitude of the friction force experienced by the front block is approximately 77.59 N.
To calculate the magnitude of the normal force exerted by the floor on the front block, we first need to understand the forces acting on the front block.
The only vertical force acting on the front block is its weight, which is equal to its mass times the acceleration due to gravity, W_front = M * g, where g ≈ 9.8 m/s² is the acceleration due to gravity.
However, there is also a vertical force from the normal force exerted by the floor on the front block, N_front. These two forces must balance each other out so that the block remains on the horizontal surface.
Since the surface is rough, there is also a horizontal frictional force acting on the front block opposing its motion.
Now, to find the magnitude of the normal force exerted by the floor on the front block, we need to consider the forces acting on the back block. The normal force exerted by the floor on the back block is equal to its weight plus the force of friction acting in the opposite direction.
Since the blocks have the same angle α and are connected together through a string or a similar constraint, the horizontal component of the forces acting on both blocks is the same.
Using the information provided, we can calculate the magnitude of the friction force experienced by the back block using the equation Friction = coefficient of kinetic friction * Normal force.
Therefore, the magnitude of the normal force exerted by the floor on the back block is N_back = W_back + Friction.
From the information given, you have already calculated the magnitude of the friction force experienced by the back block as 79.206 N. Hence, you can determine N_back = 307 N + 79.206 N = 386.206 N.
Since the horizontal component of forces acting on both blocks is the same, the magnitudes of the normal forces exerted by the floor on both blocks are equal. Hence, the magnitude of the normal force exerted by the floor on the front block is also 386.206 N.
For the magnitude of the friction force experienced by the front block, we can use the same equation as before, Friction = coefficient of kinetic friction * Normal force. Substitute the coefficient of kinetic friction as 0.258 and the normal force as 386.206 N to calculate the friction force experienced by the front block.