What force must be exerted on block A in order for block b not to fall. The coefficient of static friction between block A and B is 0.55, and the horizontal surface is frictionless.
where is A and B?
A= 100kg and B=10kg
A is on the surface and B is attached to A.
To find the force required to prevent block B from falling, we need to consider the forces acting on the system.
Let's assume the mass of block A is mA and the mass of block B is mB.
1. Identify the forces acting on block A:
- The weight of block A: We can find this by multiplying the mass of block A (mA) by the acceleration due to gravity (9.8 m/s^2). So, the weight of block A is W = mA * g, where g is 9.8 m/s^2.
- The normal force exerted on block A: Since the horizontal surface is frictionless, the normal force is equal to the weight of block A, so we can say N = W.
2. Identify the forces acting on block B:
- The normal force exerted on block B: This force is equal to the weight of block B (mB * g) because block B is not moving vertically.
- The force of static friction between block A and B: This force opposes the tendency of block B to move, and its maximum value can be calculated using the coefficient of static friction (μ) and the normal force between the two blocks (N). So, the maximum force of static friction is F_friction = μ * N.
Since we want block B not to fall, the force of static friction must be equal to or greater than the weight of block B. In other words, F_friction ≥ mB * g.
Therefore, we can set up the inequality:
μ * N ≥ mB * g
Substituting the expression for N:
μ * (mA * g) ≥ mB * g
Canceling out the acceleration due to gravity:
μ * mA ≥ mB
Finally, to find the force required to prevent block B from falling, we can multiply the mass of block B by the acceleration due to gravity:
Force_required = mB * g
So, in this case, the force required on block A must be at least equal to the weight of block B (Force_required) to prevent block B from falling.