A box of mass m=10 kg rests on a surface inclined at 0=30° to the horizontal. It is connected by a light weight cord which passes over a massless and frictionless pulley to a second box of mass Mb which hangs freely . If the coefficient of static friction is 0.40 determine what range of values for mass Mb will keep the system at rest.
first when will it slide down?
component of weight down slope = m g sin 30
normal force = m g cos 30
friction force up = 0.4 m g cos 30
other weight force up = Mb g
so for force up = force down
Mb g + 0.4 m g cos 30 = m g sin30
or Mb = 10 (sin 30 - .4 cos 30)
now
for when it slides up reverse friction effect
Mb g - 0.4 mg cos 30 = m g sin 30
or Mb = 10 (sin 30 + .4 cos 30)
this is best question
To determine the range of values for mass Mb that will keep the system at rest, we need to analyze the forces involved.
Let's start by identifying the forces acting on each box:
For the box on the inclined surface (mass m):
- Weight force (mg) acting straight downward.
- Normal force (N) acting perpendicular to the inclined surface.
- Friction force (F) acting parallel to the inclined surface.
For the hanging box (mass Mb):
- Weight force (Mbg) acting straight downward.
The system will remain at rest if the force of static friction (F) is equal to or greater than the force component of the hanging box's weight (Mbg).
The force component of the hanging box's weight parallel to the inclined surface is given by:
Mbg * sin(θ), where θ is the angle of the inclined surface (30°).
The force of static friction (F) is given by:
F = μN, where μ is the coefficient of static friction.
The normal force (N) is equal to the component of the box's weight perpendicular to the inclined surface:
N = mg * cos(θ), where θ is the angle of the inclined surface (30°).
Substituting the values into the equations, we have:
F = μN = μ * mg * cos(θ)
Mbg * sin(θ)
Setting F equal to Mbg * sin(θ), we get:
μ * mg * cos(θ) = Mbg * sin(θ)
Simplifying the equation, we get:
Mbg = (μ * mg * cos(θ)) / sin(θ)
Substituting the given values, we have:
Mbg = (0.40 * 10 kg * 9.8 m/s² * cos(30°)) / sin(30°)
Calculating this, we find:
Mbg ≈ 56.73 kg
So, the range of values for mass Mb that will keep the system at rest is from 0 kg to approximately 56.73 kg.
To determine the range of values for mass Mb that will keep the system at rest, we need to consider the forces acting on the system.
Let's label the two boxes as A (with mass m = 10 kg) and B (with mass Mb).
First, we need to analyze the forces acting on box A. Since it is resting on an inclined surface, there are two forces acting on it:
1. The gravitational force, which can be decomposed into two components: mg * sin(θ), acting parallel to the surface, and mg * cos(θ), acting perpendicular to the surface.
2. The static friction force, opposing the tendency of box A to slide down the inclined plane. This force can be calculated as fs = μs * N, where μs is the coefficient of static friction and N is the normal force. In this case, the normal force N is equal to mg * cos(θ).
Next, let's consider the forces acting on box B. It is hanging freely, so the only force acting on it is the force of gravity, which is equal to mass Mb * g.
Since there is a light cord connecting the two boxes over a pulley, the tension force in the cord is the same for both boxes.
For the system to remain at rest, the net force acting on each box must be zero, and the tension force in the cord connecting the two boxes must be equal to the force due to static friction.
So, we can write the equation for the net force on box A as:
T - fs = 0
Substituting the expressions for the tension force and static friction force:
T - (μs * mg * cos(θ)) = 0
Also, the equation for the net force on box B will be:
fs - Mb * g = 0
Substituting the expression for the static friction force:
(μs * mg * cos(θ)) - Mb * g = 0
Now, we can solve these two equations simultaneously to find the range of values for mass Mb.
Substituting the values: m = 10 kg, μs = 0.40, θ = 30°, and g = 9.8 m/s^2, we get:
(0.4 * 10 * 9.8 * cos(30°)) - Mb * 9.8 = 0
Simplifying the equation:
3.92 - 9.8Mb = 0
Rearranging the equation:
Mb = 3.92 / 9.8
Mb = 0.40 kg
Therefore, the range of values for mass Mb that will keep the system at rest is 0 kg ≤ Mb ≤ 0.40 kg.