Two blocks with masses m1 = 13.4 kg and m2 = 74.2 kg, shown in the figure, are free to move. The coefficient of static friction between the blocks is 0.55 but the surface beneath m2 is frictionless. What is the minimum force F required to hold m1 against m2?

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To determine the minimum force required to hold m1 against m2, we need to consider the forces acting on the blocks.

1. Identify the forces acting on m1:
- Weight force (mg1),
- Normal force (N1) exerted by m2 on m1,
- Friction force (f1) between m1 and m2.

2. Identify the forces acting on m2:
- Weight force (mg2),
- Normal force (N2) exerted by the surface on m2.

Since m1 is being held against m2, the friction force (f1) is at its maximum and equal to the product of the coefficient of static friction (μs) and the normal force (N1):

f1 = μs * N1

The normal force (N1) can be calculated by considering the equilibrium of the vertical forces on m1:

N1 + mg1 = N2

However, since m2 is on a frictionless surface, the only vertical force on m2 is its weight, and N2 is equal to mg2:

N1 + mg1 = mg2

Solving the equation above for N1, we get:

N1 = mg2 - mg1

Substituting back into the equation for f1:

f1 = μs * (mg2 - mg1)

Finally, the minimum force required to hold m1 against m2 is equal in magnitude but opposite in direction to the friction force (f1):

F = -f1 = -μs * (mg2 - mg1)

Plugging in the given values for masses (m1 = 13.4 kg, m2 = 74.2 kg) and the coefficient of static friction (μs = 0.55), we can compute the value of F.