Two blocks with masses m1 = 9.1 kg and m2 = 62.3 kg, shown in the figure, are free to move. The coefficient of static friction between the blocks is 0.53 but the surface beneath m2 is frictionless. What is the minimum force F required to hold m1 against m2?
To find the minimum force F required to hold m1 against m2, we need to consider the forces acting on both blocks.
For block m1 (9.1 kg):
- The force of gravity acting downwards (Fg1) = m1 * g, where g is the acceleration due to gravity (approximately 9.8 m/s^2).
For block m2 (62.3 kg):
- The force of gravity acting downwards (Fg2) = m2 * g.
- The force of static friction (Fs) acting upwards between m1 and m2. The magnitude of Fs can be given by Fs = μs * N, where μs is the coefficient of static friction and N is the normal force between m1 and m2. Since m2 is on a frictionless surface, the normal force N is equal to the force of gravity acting downwards (Fg2). Therefore, Fs = μs * Fg2.
To hold m1 against m2, the minimum force F should overcome the static friction Fs.
Since the normal force N is equal to Fg2, we can replace N in the formula for Fs and write:
Fs = μs * Fg2 = μs * (m2 * g)
Setting Fs equal to F, we have:
F = μs * (m2 * g)
Substituting the given values:
F = 0.53 * (62.3 kg * 9.8 m/s^2)
Calculating:
F ≈ 314 N
Therefore, the minimum force F required to hold m1 against m2 is approximately 314 N.