A 2.5 kg block sits on an inclined plane with a 30 degree inclination. A light cord attached to the block passes up over a light frictionless pulley at the top of the plane and is tied to a second 2.5 kg mass freely hanging vertically. The coefficients of static and kinetic friction between the block and the plane are 0.5 and 0.3. When released from rest find: the acceleration of the blocks, the tension in the string, Explain why the tension supporting the hanging block is not equal to its weight and find the time for the block on the inclined plane to travel 0.5m up the plane. And find the minimum angle of inclination at which the block on the plane will remain at rest.
Coefficient of the kinetic friction
μ₁ =0.3
Coefficient of the static friction
μ ₂ =0.5
m₁= 2.5 kg, m₂=2.5 kg
α=30°
(A)
m₁a= T- m₁gsinα- F(fr)… (1)
0=N-m₁gcosα………….(2)
m₂a= m₂g- T …………….(3)
F(fr)=μ₁N=μ₁•m₁gcosα
sum of (1) and (3):
a(m₁+m₂)=T- m₁gsinα -F(fr)- T+ m₂g=
= m₂g- m₁gsinα - μ₁m₁gcosα .
a=g(m₂-m₁sinα - μ₁m₁cosα)/(m₁+m₂) =…
T= m₂(g-a)
(B)
0= T- m₁gsinα- F(fr)… (1)
0=N-m₁gcosα………….(2)
0= m₂g - T …………….(3)
F(fr)=μ₂N= μ₂•m₁gcosα
(1) +(3)
0= T- m₁gsinα- F(fr)+ m₂g- T =
= m₂g- m₁gsinα- μ₂•m₁gcosα
m₁=m₂ =>
μ₂•cosα =1-sinα
Solve for α
To find the answers to these questions, we need to analyze the forces acting on the system and apply Newton's laws of motion. I'll explain the process step by step.
Step 1: Draw a Free Body Diagram (FBD) for each block:
- For the block on the inclined plane:
- The weight (mg) acts straight down.
- The normal force (N) acts perpendicular to the plane.
- The static friction force (fs) opposes the motion and acts parallel to the inclined plane.
- For the hanging block:
- The weight (mg) acts straight down.
- The tension in the string (T) acts upwards.
Step 2: Calculate the weights of the blocks:
- Weight (mg) = mass (m) * acceleration due to gravity (g)
- Since both blocks have a mass of 2.5 kg, the weights are:
- Weight of each block = 2.5 kg * 9.8 m/s^2 = 24.5 N
Step 3: Determine the maximum static friction force:
- The maximum static friction force (fs,max) is given by: fs,max = coefficient of static friction * normal force
- The normal force (N) is equal to the component of weight perpendicular to the plane, which is mg * cos(theta), where theta is the inclination angle (30 degrees).
- Therefore, fs,max = 0.5 * (2.5 kg * 9.8 m/s^2 * cos(30)) = 11.4 N
Step 4: Analyze the forces on the block on the inclined plane:
- The forces acting on the block on the inclined plane are: weight (24.5 N), normal force (N), static friction force (fs), and the tension in the string (T).
- The force component parallel to the inclined plane is given by: force_parallel = fs - T
- The force component perpendicular to the inclined plane is given by: force_perpendicular = N - mg * sin(theta)
Step 5: Apply Newton's Second Law to the block on the inclined plane:
- The acceleration of the block on the inclined plane (a) is in the direction of force_parallel.
- According to Newton's Second Law: force_parallel = mass * acceleration
- Therefore, fs - T = (2.5 kg) * a
Step 6: Analyze the forces on the hanging block:
- The forces acting on the hanging block are: weight (24.5 N) and tension in the string (T).
Step 7: Apply Newton's Second Law to the hanging block:
- The hanging block is freely hanging, so its acceleration is in the direction of force, which is the weight.
- According to Newton's Second Law: T - mg = (2.5 kg) * a_hanging
Step 8: Solve the equations simultaneously:
- We have two equations:
1) fs - T = (2.5 kg) * a
2) T - mg = (2.5 kg) * a_hanging
- Substitute the value of fs into equation 1 from step 3: fs = 11.4 N
- Substituting the weights of the blocks: mg = 24.5 N
- Solve the equations simultaneously to find the values of T and a.
Step 9: Determine the time for the block on the inclined plane to travel 0.5 m up the plane:
- To find the time, we need to calculate the distance traveled by the block on the inclined plane.
- The distance traveled is given by: distance = initial velocity * time + (1/2) * acceleration * time^2
- Assume the block starts from rest (initial velocity = 0) and distances traveled up and down the plane are equal.
- Use the equation above to find the value of time when distance = 0.5 m.
Step 10: Find the minimum angle of inclination for the block to remain at rest:
- For the block to remain at rest, the static friction force must be equal to or greater than the force component parallel to the inclined plane.
- Set the static friction force fs to its maximum value fs,max and solve the equation fs,max = T for the minimum angle of inclination.
By following these steps, you should be able to find the answers to the questions and understand the process behind it.