A ramp, a pulley, and two boxes.
Box A, of mass 10.0 kg, rests on a surface inclined at 37° to the horizontal. It is connected by a lightweight cord, which passes over a massless and frictionless pulley, to a second box B, which hangs freely as shown. If the coefficient of static friction is 0.40, determine what range of values for mass B will keep the system at rest.
To determine the range of values for mass B that will keep the system at rest, we need to analyze the forces acting on the system.
First, let's consider the forces acting on box A. We have the gravitational force (weight) acting vertically downward, which can be broken down into two components: one parallel to the inclined surface (mg sinθ) and one perpendicular to the surface (mg cosθ). The frictional force (static friction) acts in the opposite direction of motion, parallel to the inclined surface.
The formula for the maximum static friction force (Ff) can be calculated using the coefficient of static friction (μs) and the normal force (N), which is the perpendicular component of the weight (mg cosθ):
Ff = μsN = μs(mg cosθ)
Since the system is at rest, the maximum static friction force must be equal to (or greater than) the parallel component of weight. Therefore, we can set up an inequality:
Ff ≥ mg sinθ
Substituting the expressions for Ff and N, we have:
μs(mg cosθ) ≥ mg sinθ
Next, let's consider the forces acting on box B. We have the weight (mg) acting vertically downward, and the tension in the cord (T) acting vertically upward.
Since the system is at rest, the tension in the cord (T) must be equal to the weight of box B (mg):
T = mg
Now, using the above equation for T, we can write an equation balancing the forces acting vertically on box B:
T = mg = N - T
Simplifying the equation:
2T = N
Substituting the expression for N, we have:
2T = mg cosθ
Now we can substitute the expression for T from above:
2mg = mg cosθ
Simplifying the equation gives:
2 = cosθ
So, the angle θ is equal to 60 degrees.
Now, let's go back to the inequality we obtained earlier:
μs(mg cosθ) ≥ mg sinθ
Substituting the value of θ we just found, we have:
μs(mg cos60) ≥ mg sin60
Simplifying the equation gives:
μs(mg/2) ≥ (mg√3)/2
Now, we can cancel out the common terms of mg/2:
μs ≥ √3/2
Given that the coefficient of static friction (μs) is 0.40, we can compare it to the minimum value (√3/2) to determine the range of values for mass B. Therefore:
0.40 ≥ √3/2
Squaring both sides of the equation to eliminate the square root:
0.16 ≥ 3/4
Multiplying both sides by 4:
0.64 ≥ 3
So, the inequality is not true, which means there is no range of values for mass B that will keep the system at rest.