Two blocks, stacked one on top of the other, slide on a frictionless horizontal surface. The surface between the two blocks is rough, however, with a coefficient of static friction equal to 0.50. The top block has a mass of 2.6 kg, and the bottom block's mass is 4.0 kg. If a horizontal force F is applied to the bottom block, what is the maximum value F can have before the top block begins to slip?
To find the maximum value of force F before the top block begins to slip, we need to consider the forces acting on both blocks.
First, let's consider the forces acting on the top block. Since it is not sliding yet, the static friction force between the two blocks must be greater than or equal to the force applied on it. In other words, the static friction force must counterbalance the horizontal force applied on the top block.
The formula for static friction force is given by:
F_friction = μ_s * N
Where:
- F_friction is the static friction force
- μ_s is the coefficient of static friction
- N is the normal force between the blocks
The normal force is equal to the weight of the top block, which can be calculated as:
N = m_top * g
Where:
- m_top is the mass of the top block
- g is the acceleration due to gravity (approximately 9.8 m/s^2)
Now, let's calculate the maximum frictional force that can be exerted by the rough surface:
F_friction_max = μ_s * N
Substituting the values into the equation:
F_friction_max = 0.50 * (2.6 kg * 9.8 m/s^2)
Finally, we have:
F_friction_max = 12.74 N
Therefore, the maximum value of force F before the top block begins to slip is 12.74 N.