A man whose mass is 80kg stands on a spring weighing machine inside an elevator. What is the readingof the weighing machine when the elevator starts to ascend with an acceleration of 2ms seque. the elevator is moving with a uniform velocity, the elevator is coming to rest with a retardation 4m/s seque.
see other post on the acceleration going up. for the slowing down, use the -sign
I need answer
To determine the reading on the weighing machine at different stages, we need to consider the forces acting on the man in each scenario.
1. When the elevator starts to ascend with an acceleration of 2 m/s^2:
The forces acting on the man in this case are:
- Weight force (mg)
- Normal force (N) exerted by the spring weighing machine
- Acceleration force (ma) due to the elevator's upward acceleration
The net force acting on the man is the sum of these forces:
Net force = Weight force + Normal force + Acceleration force
Net force = mg + N + ma
Since the man is in equilibrium (not moving vertically), the net force will be zero:
mg + N + ma = 0
To find the Normal force (N), we need to isolate it:
N = -mg - ma
The reading on the weighing machine is equal to the Normal force, so the reading is:
Reading = -mg - ma
Substituting the given values:
Reading = -(80 kg)(9.8 m/s^2) - (80 kg)(2 m/s^2)
Reading = -784 N - 160 N
Reading = -944 N
Therefore, the reading on the weighing machine when the elevator starts to ascend is -944 N.
2. When the elevator is moving with a uniform velocity:
When the elevator is moving with a uniform velocity, the acceleration is zero, meaning the net force is also zero. In this case, the normal force will be equal to the weight force:
Normal force (reading) = Weight force
Reading = mg
Substituting the given values:
Reading = (80 kg)(9.8 m/s^2)
Reading = 784 N
Therefore, the reading on the weighing machine when the elevator is moving with a uniform velocity is 784 N.
3. When the elevator is coming to rest with a retardation of 4 m/s^2:
The forces acting on the man in this case are:
- Weight force (mg)
- Normal force (N) exerted by the spring weighing machine
- Retardation force (-ma) due to the elevator's downward retardation
The net force acting on the man is the sum of these forces:
Net force = Weight force + Normal force - Retardation force
Net force = mg + N - ma
Since the man is in equilibrium, the net force will be zero:
mg + N - ma = 0
To find the Normal force (N), we need to isolate it:
N = ma - mg
The reading on the weighing machine is equal to the Normal force, so:
Reading = N = ma - mg
Substituting the given values:
Reading = (80 kg)(4 m/s^2) - (80 kg)(9.8 m/s^2)
Reading = 320 N - 784 N
Reading = -464 N
Therefore, the reading on the weighing machine when the elevator is coming to rest with a retardation of 4 m/s^2 is -464 N.
To find the reading of the weighing machine, we need to consider the forces acting on the man inside the elevator in each scenario.
1. When the elevator starts to ascend with an acceleration of 2 m/s^2:
- In this case, the man experiences the force of gravity and the upward force exerted by the weighing machine.
- The net force acting on the man is the difference between these two forces.
- The net force is given by F_net = m * a, where m is the mass of the man (80 kg) and a is the acceleration of the elevator (2 m/s^2).
The net force is:
F_net = 80 kg * 2 m/s^2
F_net = 160 N
However, we need to consider that the weighing machine also applies an equal and opposite force to the man's weight (due to Newton's Third Law of Motion).
So, the reading of the weighing machine in this case will be the weight of the man plus the force due to the acceleration:
Reading of the weighing machine = Weight of the man + Force due to acceleration
= m * g + F_net
Where g is the acceleration due to gravity (~9.8 m/s^2).
Reading of the weighing machine = 80 kg * 9.8 m/s^2 + 160 N
Reading of the weighing machine ≈ 784 N + 160 N
Reading of the weighing machine ≈ 944 N
2. When the elevator is moving with a uniform velocity:
- In this case, the elevator is moving at a constant velocity, so there is no acceleration.
- The man only experiences his weight due to gravity.
- Therefore, the reading of the weighing machine will be equal to the man's weight:
Reading of the weighing machine = Weight of the man
= m * g
Reading of the weighing machine = 80 kg * 9.8 m/s^2
Reading of the weighing machine ≈ 784 N
3. When the elevator is coming to rest with a retardation of 4 m/s^2:
- In this case, the man experiences the force of gravity and the upward force exerted by the weighing machine.
- The net force acting on the man is the difference between these two forces.
- The net force is given by F_net = m * a, where m is the mass of the man (80 kg) and a is the retardation of the elevator (-4 m/s^2).
The net force is:
F_net = 80 kg * (-4 m/s^2)
F_net = -320 N (negative because it's in the opposite direction of weight)
Similarly to the first case, the reading of the weighing machine in this case will be the weight of the man plus the force due to the retardation:
Reading of the weighing machine = Weight of the man + Force due to retardation
= m * g + F_net
Reading of the weighing machine = 80 kg * 9.8 m/s^2 - 320 N
Reading of the weighing machine ≈ 784 N - 320 N
Reading of the weighing machine ≈ 464 N
So, the readings of the weighing machine in the three scenarios are as follows:
1. When the elevator starts to ascend: approximately 944 N
2. When the elevator is moving with a uniform velocity: approximately 784 N
3. When the elevator is coming to rest: approximately 464 N