Grade 11 physics
Assignment 5 Newton’s Laws
An elevator has a mass of 2250 kg.
a) What is the tension in the cable that supports the elevator when it is at rest? Complete the FBD
b) What is the tension in the cable when the elevator is moving up with an acceleration of 0.55 m/s2? Complete the FBD.
c) What is the tension in the cable when the elevator is moving down with an acceleration of 0.55 m/s2? Complete the FBD.
a) When the elevator is at rest, the net force acting on it is equal to zero. Therefore, the tension in the cable should be equal to the weight of the elevator.
The weight of the elevator can be calculated using the formula:
Weight = mass * gravity
where mass is the mass of the elevator and gravity is the acceleration due to gravity. Assuming gravity is approximately 9.8 m/s^2, we have:
Weight = 2250 kg * 9.8 m/s^2 = 22050 N
Therefore, the tension in the cable when the elevator is at rest is 22050 N.
b) When the elevator is moving up with an acceleration of 0.55 m/s^2, the net force acting on it is the sum of the tension in the cable and the force due to the elevator's acceleration. We can represent this on a Free Body Diagram (FBD) as follows:
- Tension in the cable (upward)
- Weight of the elevator (downward)
- Force due to acceleration (upward)
The net force can be calculated using the formula:
Net force = mass * acceleration
where mass is the mass of the elevator and acceleration is the acceleration of the elevator. We have:
Net force = 2250 kg * 0.55 m/s^2 = 1237.5 N
Since the elevator is moving up, the tension in the cable is greater than the weight of the elevator. Therefore, we can write:
Tension in the cable - Weight of the elevator = Net force
Tension in the cable - 22050 N = 1237.5 N
Tension in the cable = 1237.5 N + 22050 N
Tension in the cable = 23287.5 N
Therefore, the tension in the cable when the elevator is moving up with an acceleration of 0.55 m/s^2 is 23287.5 N.
c) When the elevator is moving down with an acceleration of 0.55 m/s^2, the net force acting on it is the difference between the tension in the cable and the force due to the elevator's acceleration. We can represent this on a Free Body Diagram (FBD) as follows:
- Tension in the cable (downward)
- Weight of the elevator (downward)
- Force due to acceleration (upward)
The net force can be calculated using the same formula as in part b:
Net force = mass * acceleration
Net force = 2250 kg * 0.55 m/s^2 = 1237.5 N
Since the elevator is moving down, the tension in the cable is less than the weight of the elevator. Therefore, we can write:
Weight of the elevator - Tension in the cable = Net force
22050 N - Tension in the cable = 1237.5 N
Tension in the cable = 22050 N - 1237.5 N
Tension in the cable = 20812.5 N
Therefore, the tension in the cable when the elevator is moving down with an acceleration of 0.55 m/s^2 is 20812.5 N.
a) When the elevator is at rest, the tension in the cable that supports it can be calculated using Newton's second law, which states that the net force acting on an object is equal to the product of its mass and acceleration.
In this case, since the elevator is at rest, its acceleration is zero. Therefore, the net force acting on the elevator is also zero. The forces acting on the elevator are gravity pulling the elevator downward (mg) and the tension (T) in the cable pulling it upward.
Using the equation:
Net force = T - mg
where T is the tension and m is the mass of the elevator, we can set the net force equal to zero and solve for T:
0 = T - mg
T = mg
Now, substituting the given values:
mass of the elevator (m) = 2250 kg
acceleration due to gravity (g) = 9.8 m/s^2
T = (2250 kg)(9.8 m/s^2) = 22050 N
So, the tension in the cable when the elevator is at rest is 22050 N.
b) When the elevator is moving up with an acceleration of 0.55 m/s^2, the net force acting on the elevator is the sum of the tension (T) and the force due to gravity (mg).
Using the equation:
Net force = T - mg
we can set the net force equal to ma (mass multiplied by acceleration) and solve for T:
ma = T - mg
Substituting the given values:
mass of the elevator (m) = 2250 kg
acceleration (a) = 0.55 m/s^2
acceleration due to gravity (g) = 9.8 m/s^2
2250 kg * 0.55 m/s^2 = T - 2250 kg * 9.8 m/s^2
T = (2250 kg * 0.55 m/s^2) + (2250 kg * 9.8 m/s^2)
T = 1237.5 N + 22050 N
T = 23287.5 N
So, the tension in the cable when the elevator is moving up with an acceleration of 0.55 m/s^2 is 23287.5 N.
c) When the elevator is moving down with an acceleration of 0.55 m/s^2, the net force acting on the elevator is the difference between the tension (T) and the force due to gravity (mg).
Using the equation:
Net force = T - mg
we can set the net force equal to ma (mass multiplied by acceleration) and solve for T:
ma = T - mg
Substituting the given values:
mass of the elevator (m) = 2250 kg
acceleration (a) = -0.55 m/s^2 (negative sign indicates downward acceleration)
acceleration due to gravity (g) = 9.8 m/s^2
2250 kg * (-0.55 m/s^2) = T - 2250 kg * 9.8 m/s^2
T = (2250 kg * -0.55 m/s^2) + (2250 kg * 9.8 m/s^2)
T = -1237.5 N + 22050 N
T = 20812.5 N
So, the tension in the cable when the elevator is moving down with an acceleration of 0.55 m/s^2 is 20812.5 N.
To solve the problems, we need to apply Newton's second law, which states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration. In this case, the elevator is the object in question.
First, let's consider the forces acting on the elevator. When the elevator is at rest, there are two forces acting on it: the weight force (mg) pulling it downwards and the tension force in the cable (T) pulling it upwards. The net force is therefore zero since the elevator is at rest.
a) What is the tension in the cable when the elevator is at rest?
To find the tension in the cable when the elevator is at rest, we can set up an equation using Newton's second law:
ΣF = ma
Since the net force is zero (ΣF = 0) and the mass of the elevator is given as 2250 kg, we can rearrange the equation to:
0 = 2250 kg * a
Since a = 0 (the elevator is at rest), the tension force T can be found by rearranging the equation:
T = m * g
T = 2250 kg * 9.8 m/s^2
T = 22,050 N
Therefore, the tension in the cable when the elevator is at rest is 22,050 N.
b) What is the tension in the cable when the elevator is moving up with an acceleration of 0.55 m/s^2?
When the elevator is moving up with an acceleration of 0.55 m/s^2, we need to consider the additional force being exerted on the elevator. The equation remains the same:
ΣF = ma
Now, the net force is the sum of the tension force (T) acting upward and the weight force (mg) acting downward:
T - mg = ma
Substituting the given values, we can solve for T:
T = m(g + a)
T = 2250 kg * (9.8 m/s^2 + 0.55 m/s^2)
T = 23,175 N
Therefore, the tension in the cable when the elevator is moving up with an acceleration of 0.55 m/s^2 is 23,175 N.
c) What is the tension in the cable when the elevator is moving down with an acceleration of 0.55 m/s^2?
When the elevator is moving down with an acceleration of 0.55 m/s^2, the equation remains the same:
ΣF = ma
Now, the net force is the difference between the tension force (T) acting downward and the weight force (mg) acting downward:
T - mg = -ma
(-ma since the acceleration is in the opposite direction)
Rearranging the equation, we can solve for T:
T = m(g - a)
T = 2250 kg * (9.8 m/s^2 - 0.55 m/s^2)
T = 21,825 N
Therefore, the tension in the cable when the elevator is moving down with an acceleration of 0.55 m/s^2 is 21,825 N.
Remember to always draw the free-body diagram (FBD) to visualize the forces acting on the object before solving the equations.