Two blocks with masses m1 = 24.6 kg and m2 = 60.0 kg, shown in the figure, are free to move. The coefficient of static friction between the blocks is 0.67 but the surface beneath m2 is frictionless. What is the minimum force F required to hold m1 against m2?
Please describe what you see in the figure, because we don't.
To determine the minimum force required to hold m1 against m2, we need to analyze the forces acting on the system.
First, let's consider the forces acting on m1:
1. Weight (W1): This force is equal to the mass (m1) multiplied by the acceleration due to gravity (g = 9.8 m/s^2). W1 = m1 * g.
2. Normal force (N1): This is the force exerted by the surface on m1 in the upward direction. In this case, N1 is equal to W1 since the block is on a horizontal surface and not accelerating vertically.
Now, let's examine the forces acting on m2:
1. Weight (W2): This force is equal to the mass (m2) multiplied by the acceleration due to gravity. W2 = m2 * g.
2. Normal force (N2): The only force acting in the vertical direction on m2 is W2. N2 is equal to W2 since the block is on a horizontal surface and not accelerating vertically.
3. Frictional force (Ff): This force opposes the relative motion or tendency of motion between the two blocks. The coefficient of static friction (μ) is given as 0.67. The formula for the maximum static frictional force (Fsmax) is Fsmax = μ * N2. However, since the surface beneath m2 is frictionless, the frictional force on m2 is zero.
Now, to find the minimum force required to hold m1 against m2, we need to consider the maximum static frictional force between the two blocks, exerted by m1 on m2.
The maximum static frictional force (Fsmax) can be calculated using the formula Fsmax = μ * N2. In this case, N2 is equal to W2 since there are no other forces acting in the vertical direction.
Fsmax = μ * W2
= μ * (m2 * g)
= 0.67 * (60.0 kg * 9.8 m/s^2)
≈ 392.4 N
Therefore, the minimum force F required to hold m1 against m2 should be larger than the maximum static frictional force of 392.4 N.