Two blocks with masses m1 = 10.9 kg and m2 = 79.3 kg, shown in the figure, are free to move. The coefficent of static friction between the blocks is 0.66 but the surface beneath m2 is frictionless. What is the minimum force F required to hold m1 against m2?
To calculate the minimum force required to hold m1 against m2, we need to analyze the forces acting on the system.
First, let's consider m1:
1. The weight of m1, W1 = m1 * g, where g is the acceleration due to gravity (approximately 9.8 m/s^2).
2. The normal force acting on m1, N1, which is the force exerted by m2 on m1 in the vertical direction.
3. The static friction force acting on m1, F friction1, which is given by the equation F friction1 = μ * N1, where μ is the coefficient of static friction between the blocks.
Now, let's consider m2:
1. The weight of m2, W2 = m2 * g, where g is the acceleration due to gravity.
2. The force exerted by m1 on m2 is equal in magnitude and opposite in direction to F friction1.
Since m2 is frictionless, there is no horizontal frictional force acting on m2.
To find the minimum force F required to hold m1 against m2, we need to set up an equation based on the forces acting on m1 in the horizontal direction:
F - F friction1 = 0
Substituting the equation for F friction1, we have:
F - μ * N1 = 0
Now, let's find N1:
N1 is equal to the force exerted by m2 on m1 in the vertical direction. This force is the weight of m2, W2. Therefore:
N1 = W2 = m2 * g
Substituting this into the equation above, we have:
F - μ * (m2 * g) = 0
Simplifying the equation, we can solve for F:
F = μ * (m2 * g)
Plugging in the given values:
F = 0.66 * (79.3 kg * 9.8 m/s^2)
Calculating the result:
F ≈ 515.3864 N
Therefore, the minimum force F required to hold m1 against m2 is approximately 515.3864 Newtons.