The three objects in the figure are connected by a light cord.
A triangular structure is oriented such that its base rests upon a horizontal surface, its right side is perpendicular to its base, and its left side forms an incline of angle 𝜃 with the base. A pulley is attached to the structure's apex. Three objects (masses m, 2m, and M) are connected by two cords (tensions T1 and T2) and arranged as follows
Masses 2m and m are, respectively, attached to the left and right ends of cord T1 such that mass m is nearest to the structure's apex.
Masses m and M are, respectively, attached to the left and right ends of cord T2 such that cord T2 passes over the pulley and mass M hangs freely.
(a)
If the inclined plane is frictionless, find the following, assuming the system is in equilibrium. (Use the following as necessary: m, g, and 𝜃.)
mass M =
tension T1 =
tension T2 =
(b)
If the value of M is double that from part (a), find the following. Note that the system will no longer be in equilibrium. (Use the following as necessary: m, g, and 𝜃.)
acceleration a =
tension T1 =
tension T2 =
(c)
If the coefficient of static friction between the blocks and the inclined plane is 𝜇s, and the system is in equilibrium, find the following. (Use the following as necessary: 𝜇s, m, g, and 𝜃.)
Mmax =
Mmin =
(d)
Compare the values of T2 when M has its minimum and maximum values. (Use the following as necessary: 𝜇s, 𝜃, m, and g.)
T2, max − T2, min =
(a)
mass M = 2m + m = 3m
tension T1 = mg/sin(theta)
tension T2 = Mg - T1*cos(theta)
(b)
acceleration a = (2g)/(sin(theta) + 2*cos(theta))
tension T1 = ma
tension T2 = M(g + a)
(c)
Mmax = (m*sin(theta))/(mu_s - cos(theta))
Mmin = (m*sin(theta))/(mu_s + cos(theta))
(d)
T2, max - T2, min = [(Mmax)g - (T1max)*cos(theta) - (Mmin)g + (T1min)*cos(theta)]
= [(Mmax - Mmin)g + (T1min - T1max)*cos(theta)]
(a) In equilibrium, the net force acting on the system should be zero in both the vertical and horizontal directions. Let's analyze the forces acting on each object:
For mass M:
1. Weight (mg) acting downwards.
2. Tension T2 acting upwards.
For mass 2m:
1. Weight (2mg) acting downwards.
2. Tension T1 acting to the right.
For mass m:
1. Weight (mg) acting downwards.
2. Tension T1 acting to the left.
3. Normal force (N) acting perpendicular to the inclined plane.
4. Frictional force (f = 𝜇sN) acting parallel to the inclined plane in the opposite direction to T1.
Since the system is in equilibrium, the net force acting on each object is zero:
Vertical forces:
T2 - Mg = 0 ------(1)
Horizontal forces:
T1 - T1cos𝜃 - f = 0
T1(1 - cos𝜃) - 𝜇sN = 0 ------(2)
Since the inclined plane is frictionless, 𝜇s = 0. Therefore, equation (2) can be simplified as:
T1(1 - cos𝜃) = 0
From equation (1), we can solve for T2:
T2 = Mg ------(3)
From equation (2), we can solve for T1:
T1 = 0/1-cos𝜃 = 0 (since friction is absent) ------(4)
Thus, in the case of a frictionless inclined plane, the values are:
mass M = M
tension T1 = 0
tension T2 = Mg
(b) If the value of M is doubled, the system will no longer be in equilibrium. To find the acceleration and the new tensions, let's analyze the forces again:
Vertical forces:
T2 - 2Mg = Ma ------(5)
Horizontal forces:
T1(1 - cos𝜃) - 𝜇sN = Mgsin𝜃 ------(6)
From equations (5) and (6), we can solve for the acceleration a, tension T1, and tension T2.
(c) If the coefficient of static friction between the blocks and the inclined plane is 𝜇s, the system is in equilibrium when the frictional force is at its maximum:
Frictional force (f = 𝜇sN) should be equal to the force exerted by T1:
𝜇sN = T1
Since the inclined plane is at an angle 𝜃, the normal force N can be written as:
N = Mgcos𝜃
Substituting this into the equation, we get:
𝜇sMgcos𝜃 = T1
Thus, the maximum value of M that can be in equilibrium is:
Mmax = T1 / (𝜇s * g * cos𝜃)
To find the minimum value of M, we assume that the frictional force is zero. Thus:
T1 = 0
For the system to be in equilibrium, T2 must balance the weight of masses m and 2m. So:
T2 = (m + 2m)g
Thus, the minimum value of M that can be in equilibrium is:
Mmin = (m + 2m)g
(d) Comparing the values of T2 when M has its minimum and maximum values:
When M is at its maximum value (Mmax):
T2, max = Mmax * g
When M is at its minimum value (Mmin):
T2, min = Mmin * g
The difference between T2 when M is at its maximum and minimum values is:
T2, max - T2, min = (Mmax - Mmin) * g
To solve this problem, we will break it down into different parts.
(a) If the system is in equilibrium and the inclined plane is frictionless, we can use the principles of forces and torques to find the unknowns.
1. Start by drawing a free-body diagram for each mass:
- For mass M, there are two forces acting on it: the tension T2 pulling it upwards, and its weight Mg pulling it downwards.
- For mass 2m, there are three forces acting on it: the tension T1 pulling it to the left, the tension T2 pulling it upwards, and its weight 2mg pulling it downwards.
- For mass m, there are two forces acting on it: the tension T1 pulling it to the right and its weight mg pulling it downwards.
2. Write the equilibrium equations for each mass:
- For mass M: T2 - Mg = 0 (since it is not accelerating vertically)
- For mass 2m: T2 - T1 - 2mg = 0 (since it is not accelerating vertically)
- For mass m: T1 - mg = 0 (since it is not accelerating horizontally)
3. Solve the equations simultaneously:
From equation 1: T2 = Mg
From equation 3: T1 = mg
Substitute these values into equation 2:
Mg - mg - 2mg = 0
Mg - 3mg = 0
M = 3m
So, the mass of M is three times the mass of m.
(b) If the value of M is double that from part (a), the system will no longer be in equilibrium, and there will be acceleration.
1. Repeat step 1 from part (a) to draw the free-body diagrams for each mass.
2. Modify the equilibrium equations:
- For mass M: T2 - Mg = Ma (since it is accelerating upwards)
- For mass 2m: T2 - T1 - 2mg = 2ma (since it is accelerating to the left)
- For mass m: T1 - mg = ma (since it is accelerating to the right)
3. Solve the equations simultaneously:
From equation 1: T2 = Mg + Ma
From equation 3: T1 = mg + ma
Substitute these values into equation 2:
Mg + Ma - mg - ma - 2mg = 2ma
Mg - 3mg = 4ma
M = 3m
So, the mass of M is still three times the mass of m, even if it is doubled.
(c) If the coefficient of static friction between the blocks and the inclined plane is μs, and the system is in equilibrium, we need to consider the frictional force acting on mass 2m.
1. Add the frictional force to the free-body diagram for mass 2m. The frictional force opposes the motion and acts to the right.
2. Modify the equilibrium equations for mass 2m:
- For mass 2m: T2 - T1 - 2mg - Fs = 0 (since it is not accelerating horizontally)
3. Solve the equation:
T2 - T1 - 2mg - μsN = 0
T2 - T1 - 2mg - μsmgcos𝜃 = 0
T2 - T1 = (2 + μscos𝜃)mg
So, the difference between T2 and T1 is equal to (2 + μscos𝜃)mg.
(d) To compare the values of T2 when M has its minimum and maximum values, we need to consider the effect of static friction.
1. Rewrite the equation from part (c):
T2 - T1 = (2 + μscos𝜃)mg
2. When M has its minimum value, the frictional force Fs is at its maximum to prevent mass 2m from sliding. The maximum frictional force is given by μsN, where N is the normal force.
N = mgcos𝜃 (since the incline is frictionless)
Fs = μsN = μsmgcos𝜃
Substitute this into the equation:
T2 - T1 = (2 + μscos𝜃)mg - μsmgcos𝜃
T2 - T1 = (2 + μscos𝜃 - μscos𝜃)mg
T2 - T1 = 2mg
3. When M has its maximum value, there is no frictional force since mass 2m does not require it to prevent sliding. So, we can ignore the term involving static friction in the equation:
T2 - T1 = (2 + μscos𝜃 - μscos𝜃)mg
T2 - T1 = 2mg
Therefore, the difference T2, max - T2, min = 2mg.
This means that the difference between the tensions T2 when M has its minimum and maximum values is equal to 2mg.