The two blocks (m = 37 kg and M = 109 kg) in the figure below are not attached to each other. The coefficient of static friction between the blocks is µs = 0.55, but the surface beneath the larger block is frictionless. What is the minimum magnitude of the horizontal force vector F required to keep the smaller block from slipping down the larger block?
__________ N
Can you please describe the figure?
Are the blocks one on top of the other, are they on an incline plane, at what point is the force to be applied (the small block, the large block)? etc.
To find the minimum magnitude of the horizontal force required to keep the smaller block from slipping down the larger block, we need to analyze the forces acting on the blocks.
First, let's consider the smaller block. The forces acting on it are its weight (mg) and the force of static friction (f_s) between the two blocks. Since the block is not slipping, the friction force is at its maximum value and is given by:
f_s = µs * N
where µs is the coefficient of static friction and N is the normal force between the two blocks.
The normal force N is equal to the weight of the larger block, as there are no other vertical forces acting on it. Therefore, N = Mg.
Now, let's consider the larger block. The only force acting on it in the horizontal direction is the force F. The weight of the larger block is acting vertically and does not have a horizontal component.
For the smaller block to remain stationary and not slip, the force of static friction between the blocks must be equal to or greater than the horizontal component of the force F.
Therefore, we have:
f_s ≥ F
Substituting the values of f_s and N:
µs * Mg ≥ F
Now, we can substitute the given values into the equation:
0.55 * (37 kg + 109 kg) * 9.8 m/s^2 ≥ F
Simplifying the equation:
F ≤ 645.6 N
Therefore, the minimum magnitude of the horizontal force required to keep the smaller block from slipping down the larger block is 645.6 N.