A 5kg object placed on a friction-less, horizontal table connected to a string that passes over a pulley and then is fastened to a hanging 9kg object.
(A) Draw a free body diagram of both the objects.
(B) The Coefficient of kinetic friction between the object and the surface is 0.35 and the coefficient of static friction is 0.46. Find: (i)The acceleration of the two objects
(ii)The tension in the cable
T = spring tension
WHEN NO FRICTION:
Object 1
force up from table = 5 g but irrelevant, balanced by gravity 5 g down
horizontal force = T
so
T = 5 a
Object 2
force down = 9 g
force up = T
so
9 g - T = 9 a
WITH Friction
Object 1
Horizontal force =
T - mu (5 g) = 5 a
Object 2
9 g - T = 9 a
======================
so add two equations to eliminate T
9 g - 5 mu g = 14 a
a = (g/14)(9-5 mu)
If mu < 9/5 it moves so use mu = .35 , the sliding coeffficient
a = (9.81/14)(9 - 5*.35)
now use a to go back and get
T=9(9.81-a)
Asweer
(A) Free body diagrams:
For the 5kg object:
- Weight (mg) downwards
- Tension of the cable (T) upwards
For the 9kg object:
- Weight (mg) downwards
- Tension of the cable (T) upwards
(B) (i) To find the acceleration of the two objects, we can use Newton's second law of motion: ΣF = ma, where ΣF is the sum of the forces acting on an object, m is the mass of the object, and a is the acceleration.
For the 5kg object:
- ΣF = T - mg
- Since the table is frictionless, there is no horizontal force other than T, so ΣF in the horizontal direction is T
- T - mg = ma
- T = ma + mg
For the 9kg object:
- ΣF = mg - T
- mg - T = ma
- T = mg - ma
Since both objects are connected by the same string, the tension in the cable is the same for both objects. Therefore, we can equate the tensions:
ma + mg = mg - ma
Simplifying the equation:
2ma = mg
a = g/2
a = 9.8 m/s^2 / 2
a = 4.9 m/s^2
The acceleration of the two objects is 4.9 m/s^2.
(ii) To find the tension in the cable, we can use the equation T = mg - ma (from the 9kg object's free body diagram):
T = (9kg)(9.8 m/s^2) - (9kg)(4.9 m/s^2)
T = 88.2 N - 44.1 N
T = 44.1 N
The tension in the cable is 44.1 N.
(A) To draw a free body diagram, we need to consider all the forces acting on both objects.
Free body diagram of the 5kg object:
- There is the weight (mg) acting downwards.
- There is tension in the string (T) acting upwards.
Free body diagram of the 9kg object:
- There is the weight (mg) acting downwards.
- There is tension in the string (T) acting upwards.
(B) To find the acceleration of the two objects and the tension in the cable, we need to analyze the forces acting on them.
For the 5kg object:
- The net force acting on it is the difference between the tension force and the force of friction. Since the table is friction-less, there is no friction force.
- Therefore, the net upward force is the tension force (T).
For the 9kg object:
- The net force acting on it is the difference between the tension force and its weight. The weight is given by mg.
- Therefore, the net upward force is T - mg.
Using Newton's second law of motion, the net force is equal to the mass multiplied by the acceleration:
- For the 5kg object, the net force is T, so T = m1 * a (where m1 = mass of the 5kg object).
- For the 9kg object, the net force is T - mg, so T - mg = m2 * a (where m2 = mass of the 9kg object).
Given the masses of the objects (m1 = 5kg, m2 = 9kg) and the acceleration (a), we can solve these equations simultaneously to find the values of acceleration and tension.
To find the coefficient of kinetic friction:
- The coefficient of kinetic friction is used to find the force of friction acting on the object.
- The force of friction is given by the equation: f = μ *N, where μ is the coefficient of friction and N is the normal force.
- In this case, since the object is on a horizontal table, the normal force is equal to the weight of the object (mg).
- Therefore, the force of friction is given by f = μ * mg.
- The force of friction acts in the opposite direction to the applied force (tension). So, it can be written as -μ * mg.
- This force should be in equilibrium with the applied force, so we can set T - μ * mg = m2 * a.
Solving the equations T = m1 * a and T - μ * mg = m2 * a will give us the values of acceleration (a) and tension (T).
(i) The acceleration of the two objects can be found by solving the equations T = m1 * a and T - μ * mg = m2 * a simultaneously.
(ii) The tension in the cable can be found by substituting the value of acceleration (a) into the equation T = m1 * a or T - μ * mg = m2 * a.