Two blocks with masses m1 = 24.3 kg and m2 = 66.9 kg, shown in the figure, are free to move. The coefficient of static friction between the blocks is 0.34 but the surface beneath m2 is frictionless. What is the minimum force F required to hold m1 against m2
Well, according to the figure, it seems like m1 and m2 aren't the best of friends. They want to slide away from each other, but we need to keep them together. The coefficient of static friction tells us that they can resist sliding, but only up to a certain point.
Now, looking at m2, it's got a frictionless surface beneath it. It's like it's at a fancy party, sliding around effortlessly. Lucky m2!
To keep m1 from saying, "See ya, m2!" we need to apply a force that counteracts the force of friction between them. The maximum force of static friction (fs) between them can be found using the equation fs = µs * N, where µs is the coefficient of static friction, and N is the normal force.
Since the surface beneath m2 is frictionless, the normal force acting on m1 is simply its weight. Therefore, N = m1 * g, where g is the acceleration due to gravity.
So, fs = 0.34 * (m1 * g).
Now, we need to apply a force F that's greater than fs to keep the two blocks together. The minimum force required would be just slightly greater than fs. So, the minimum force F would be fs + ε, where ε is a small amount more than fs.
Therefore, F = fs + ε = 0.34 * (m1 * g) + ε.
Just remember that ε is small but necessary so that m1 doesn't go on an adventure without m2.
To determine the minimum force (F) required to hold m1 against m2, we first need to consider the forces acting on each block.
For block m1:
- The force of gravity acting on m1 is given by Fg1 = m1 * g (where g is the acceleration due to gravity).
- The frictional force between m1 and m2 is given by Ff1 = μs * N1 (where μs is the coefficient of static friction and N1 is the normal force acting on m1).
For block m2:
- The force of gravity acting on m2 is given by Fg2 = m2 * g.
- There is no frictional force acting on m2 since the surface beneath it is frictionless.
To hold m1 against m2, the force of friction (Ff1) must be equal to or greater than the force trying to pull m1 away (Fg2). Therefore, we can set up the following inequality:
Ff1 ≥ Fg2
Substituting the equations for Ff1 and Fg2, we get:
μs * N1 ≥ m2 * g
Now, let's determine the normal force N1 acting on m1. Since the two blocks are in contact, the normal force between them is equal in magnitude and opposite in direction. Hence:
N1 = N2
Next, we can calculate the normal force N2 acting on m2:
N2 = m2 * g
Now, the inequality becomes:
μs * N2 ≥ m2 * g
Substituting the value of N2, we get:
μs * m2 * g ≥ m2 * g
Simplifying the equation, we find:
μs * m2 ≥ m2
Dividing both sides of the equation by m2 (which is positive), we get:
μs ≥ 1
Since the coefficient of static friction cannot be greater than 1, we can conclude that the minimum force (F) required to hold m1 against m2 is when Ff1 is equal to Fg2, which occurs when μs = 1.
Therefore, if the coefficient of static friction is 1 or greater, any force (F) greater than or equal to Fg2 will hold m1 against m2.