Two blocks of masses 40.0kg and 20.0kg are stacked on a table with the heavier block on top. The coefficient of static friction is 0.600 between the top and bottom blocks and 0.300 between the bottom block and the table. What is the required magnitude of a horizontal force that, if applied to the top block, would move the blocks with the same net acceleration?

You want the force greater than the force of friction at the table surface, but not greater than the friction between the two blocks.

they're asking the same questions in 2016. lol

haha yeah, was looking for the answer, still no update

To find the required magnitude of the horizontal force that would move the blocks with the same net acceleration, we need to consider the forces acting on both blocks.

1. Calculate the maximum static friction force between the top and bottom blocks:
The maximum static friction force is given by the formula: F_friction = coefficient_of_friction * normal_force.
The normal force between the two blocks is equal to the weight of the top block, which is equal to the mass of the top block multiplied by the acceleration due to gravity (9.8 m/s^2).

For the top and bottom blocks, the normal force is the same, as the bottom block's weight is supporting the weight of the top block.
Therefore, the maximum static friction force between the two blocks can be calculated as:

F_friction_between_blocks = coefficient_of_static_friction * (mass_of_top_block + mass_of_bottom_block) * gravitational_acceleration.

Substituting the given values:
F_friction_between_blocks = 0.600 * (40.0 kg + 20.0 kg) * 9.8 m/s^2.

2. Calculate the maximum static friction force between the bottom block and the table:
The maximum static friction force between the bottom block and the table can be calculated using the same formula:
F_friction_on_table = coefficient_of_static_friction * mass_of_bottom_block * gravitational_acceleration.

Substituting the given values:
F_friction_on_table = 0.300 * 20.0 kg * 9.8 m/s^2.

3. Determine the required magnitude of the horizontal force:
The required force should be greater than the force of friction at the table surface (F_friction_on_table) but not greater than the friction between the two blocks (F_friction_between_blocks). Therefore, the required magnitude of the horizontal force should be in the range between F_friction_on_table and F_friction_between_blocks.

Therefore, the required magnitude of the horizontal force is between:
F_friction_on_table = 0.300 * 20.0 kg * 9.8 m/s^2
F_friction_between_blocks = 0.600 * (40.0 kg + 20.0 kg) * 9.8 m/s^2.