Question
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?
Answers
We can start by calculating the maximum static friction force between the two blocks. This can be found using the equation:
f_max = μ * m_top * g
where f_max is the maximum static friction force, μ is the coefficient of static friction, m_top is the mass of the top block, and g is the acceleration due to gravity (9.81 m/s²).
f_max = 0.50 * 2.6 kg * 9.81 m/s²
f_max = 12.77 N
Now, we know that as long as the force F remains less than or equal to this maximum static friction force, the top block will not slip. Therefore, the maximum value F can have before the top block begins to slip is 12.77 N.
f_max = μ * m_top * g
where f_max is the maximum static friction force, μ is the coefficient of static friction, m_top is the mass of the top block, and g is the acceleration due to gravity (9.81 m/s²).
f_max = 0.50 * 2.6 kg * 9.81 m/s²
f_max = 12.77 N
Now, we know that as long as the force F remains less than or equal to this maximum static friction force, the top block will not slip. Therefore, the maximum value F can have before the top block begins to slip is 12.77 N.
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