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.
1. Draw a free-body diagram for the front block. We have the weight (W) acting vertically downward, the normal force (N) acting vertically upward, and the friction force (f) acting horizontally in the opposite direction of motion.
2. Since the block is on a horizontal surface with no vertical acceleration, the net force in the vertical direction is zero. Therefore, the sum of the vertical forces must be equal to zero: N - W = 0.
3. The weight (W) can be calculated as the mass (M) of the front block multiplied by the acceleration due to gravity (g = 9.8 m/s²): W = M * g.
W = 30.7 kg * 9.8 m/s² = 300.86 N (rounded to three decimal places).
4. Using the equation from step 2, we have N - 300.86 N = 0. Solving for N, we find:
N = 300.86 N.
So, the magnitude of the normal force exerted by the floor on the front block is 300.86 N.
Now let's calculate the magnitude of the friction force experienced by the front block.
5. The friction force can be calculated using the equation: f = μN, where μ is the coefficient of kinetic friction and N is the normal force.
6. Substituting the values we already calculated, we have:
f = 0.258 * 300.86 N ≈ 77.707 N (rounded to three decimal places).
Therefore, the magnitude of the friction force experienced by the front block is approximately 77.707 N.