Determine the elevator's acceleration

• y90x
In summary: Since the man isn’t part of the system , his mass or weigh shouldn’t be included... or is it?No, the man's mass or weight should not be included in the equation.
y90x

Homework Statement

An 82 kg man inside a 40-kg dumb-waiter pulls down on the rope. At that moment the scale on which he is standing reads 209 N. Determine the elevator's acceleration.

F=ma

The Attempt at a Solution

I know the acceleration is 0.152 m/s^2 upwards , I need help solving the problem
https://www.physicsforums.com/attachments/215547
I drew a diagram , I’m no sure I am correct because I keep getting a number greater

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Draw a free body diagram, then apply the relevant equation that you posted.

kuruman said:
Draw a free body diagram, then apply the relevant equation that you posted.

Would the rope have an impact though?

y90x said:
Would the rope have an impact though?
What do you think? Would the man be able to raise himself without it?

kuruman said:
What do you think? Would the man be able to raise himself without it?

Fg=mg
=(82kg)(9.8) = 803.6 N

And the force exerted on the rope is T
So the formula would be,

(Fn +T)-Fg=ma

Am I wrong or missing something ?

y90x said:
missing something ?
That's just for the man, right?
What about the equation for the dumb waiter? Make sure to use a different variable for the mass.

You have the equation with the man as the system. However the tension is unknown. How are you going to find that?

kuruman said:
You have the equation with the man as the system. However the tension is unknown. How are you going to find that?

I’m no sure , I have two missing variables. Wouldn’t that mean to supplement a (acceleration) with F/M ?

y90x said:
I’m no sure , I have two missing variables. Wouldn’t that mean to supplement a (acceleration) with F/M ?
See post #6.

haruspex said:
That's just for the man, right?
What about the equation for the dumb waiter? Make sure to use a different variable for the mass.

Would this be all the forces ? Or would the dumb waiter exert a normal force as well

Attachments

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y90x said:
View attachment 215554
Would this be all the forces ? Or would the dumb waiter exert a normal force as well
You have ΣF=ma equation for the man. Your second one can be either for the dumb waiter only or for the man+dumb waiter combination. Your choice.
If for the dumb waiter, what force does the man exert on it? Remember Newton's laws.
If for the combination, ignore forces between the man and the dumb waiter since they are internal to the system, but be careful about the force the rope exerts on the combination.

haruspex said:
You have ΣF=ma equation for the man. Your second one can be either for the dumb waiter only or for the man+dumb waiter combination. Your choice.
If for the dumb waiter, what force does the man exert on it? Remember Newton's laws.
If for the combination, ignore forces between the man and the dumb waiter since they are internal to the system, but be careful about the force the rope exerts on the combination.

If I were to chose only the dumb waiter c wouldn’t it be the gravitational force going down and the tension ?
So it’ll be
Fnet=ma
T-Md=Md(a)

y90x said:
If I were to chose only the dumb waiter c wouldn’t it be the gravitational force going down and the tension ?
So it’ll be
Fnet=ma
T-Md=Md(a)
You left out g, and another force. As I wrote, remember Newton's laws.

haruspex said:
You left out g, and another force. As I wrote, remember Newton's laws.

Oh yes !
Wouldn’t it be the man’s force being exerted down on the elevator ?

T- Md•g - M•g = a(Md+M)

y90x said:
T- Md•g - M•g = a(Md+M)
In Fnet = ma, m is the mass of the system. Your system is the dumbwaiter, so what quantity should multiply a?
Also on the left side you have M⋅g. That's the force that the Earth exerts on the man, however neither the Earth nor the man are part of this system. The force exerted by the man on the dumbwaiter is not his weight. What is that force? As @haruspex wrote, remember Newton's Laws.

kuruman said:
In Fnet = ma, m is the mass of the system. Your system is the dumbwaiter, so what quantity should multiply a?
Also on the left side you have M⋅g. That's the force that the Earth exerts on the man, however neither the Earth nor the man are part of this system. The force exerted by the man on the dumbwaiter is not his weight. What is that force? As @haruspex wrote, remember Newton's Laws.

So you would multiply a by 40? The mass of the dumbwaiter . So then the missing force would be the normal force in which the scale reads ?

T - Mdg + Fn = Mda ?
Since the man isn’t part of the system , his mass or weigh shouldn’t be included ?

y90x said:
T - Mdg + Fn = Mda ?
Which way is the normal force acting on the dumb waiter?

y90x said:
Oh yes !
Wouldn’t it be the man’s force being exerted down on the elevator ?
Yes.
T- Md•g - M•g = a(Md+M)
However, the man’s force being exerted down on the elevator isn't mg in this case.

What does Newton's third law say about the force the man exerts on the elevator as compared with the force the elevator exerts on the man?

haruspex said:
Which way is the normal force acting on the dumb waiter?

It’s going up , so shouldn’t it be added to tension

SammyS said:
Yes.

However, the man’s force being exerted down on the elevator isn't mg in this case.

What does Newton's third law say about the force the man exerts on the elevator as compared with the force the elevator exerts on the man?

That they should be equal but since the elevator is accelerating wouldn’t that make his weight seen heavier or lighter depending the speed it’s going ?

y90x said:
That they should be equal but since the elevator is accelerating wouldn’t that make his weight seen heavier or lighter depending the speed it’s going ?
No. The weight is the force Mmang with which the Earth attracts the man. What changes is the normal force exerted by the scale on the man that always adapts itself to provide the observed acceleration. The scale measures the normal exerted on its surface. Just push down on a scale with your hand to convince yourself that it is not displaying your weight.

kuruman said:
No. The weight is the force Mmang with which the Earth attracts the man. What changes is the normal force exerted by the scale on the man that always adapts itself to provide the observed acceleration. The scale measures the normal exerted on its surface. Just push down on a scale with your hand to convince yourself that it is not displaying your weight.

Oh, I see
So then it’ll be
T-209N-40g=40a ?

y90x said:
So then it’ll be
T-209N-40g=40a ?
It'll be that.

SammyS
kuruman said:
It'll be that.

But tension is still unknown, to solve it would it be

T - 82g = 209N because the elevator is going upwards, or
82g-T =209N

You have two equations, one for the man and one for the dumbwaiter, and two unknowns, the acceleration and the tension.

kuruman said:
You have two equations, one for the man and one for the dumbwaiter, and two unknowns, the acceleration and the tension.

So how do you find tension ?

y90x said:
It’s going up , so shouldn’t it be added to tension
The normal force between the man and the dumb waiter (we can ignore the scale since that is considered massless, so just transmits the force straight through) must be up on one and down on the other. Does it act down on the man and up on the dumb waiter or the other way around?

haruspex said:
The normal force between the man and the dumb waiter (we can ignore the scale since that is considered massless, so just transmits the force straight through) must be up on one and down on the other. Does it act down on the man and up on the dumb waiter or the other way around?

Wouldn’t it be the other way around ? Up on the man and down on the dumbwaiter ?

Up on the man and down on the elevator is correct. Your equation in post #22 is also correct. Where does your equation in post #24 come from?

kuruman said:
Up on the man and down on the elevator is correct. Your equation in post #22 is also correct. Where does your equation in post #24 come from?

My teacher explained to use two different equations when two variables are unknown, I was trying to find another equation to substitute T with

y90x said:
My teacher explained to use two different equations when two variables are unknown, I was trying to find another equation to substitute T with
You have the correct information for the man in post #5.

y90x said:
Fg=mg
=(82kg)(9.8) = 803.6 N

And the force exerted on the rope is T
So the formula would be,

(Fn +T)-Fg=ma

Am I wrong or missing something ?
Use 209 Newtons for Fn and mg for Fg, of course with 82 kg for the man's mass.

Then as @kuruman wrote, your equation in post #22 is correct.
y90x said:
Oh, I see
So then it’ll be
T-209N-40g=40a ?
Use those two equations to eliminate T (or eliminate a if you'd rather find the tension).

y90x said:
Wouldn’t it be the other way around ? Up on the man and down on the dumbwaiter ?
Right, but in post #16 you had an equation for the dumb waiter with T...+Fn = ...
If the tension is acting upwards on the dumb waiter and the normal force is acting downwards then they cannot both be positive on the same side of tne equation.

I have a suggestion for OP, @y90x. Why don't you write two equations for Newton's 2nd Law, Fnet = ma, one for which the system is the man and one for which the system is the dumbwaiter and in which only symbols appear, not numbers, and post them. One (or more) of us will review these equations and point out what is wrong (if anything) with them. At this point, it seems to me there is some tail-chasing going on here so a fresh start would be beneficial.

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SammyS

1. What is the definition of acceleration?

Acceleration is the rate of change of an object's velocity over time. It is a vector quantity, meaning it has both magnitude and direction.

2. How is acceleration calculated?

Acceleration is calculated by dividing the change in an object's velocity by the time it takes for that change to occur. The formula for acceleration is a = (vf - vi) / t, where a is acceleration, vf is final velocity, vi is initial velocity, and t is time.

3. How does an elevator's acceleration affect its passengers?

An elevator's acceleration affects its passengers by causing a sensation of weightlessness or heaviness. When the elevator is accelerating upwards, passengers will feel heavier due to the increased normal force from the elevator floor. When the elevator is accelerating downwards, passengers will feel lighter due to the decreased normal force.

4. What factors can affect an elevator's acceleration?

The main factors that can affect an elevator's acceleration are the weight of the elevator and its passengers, the strength of the elevator's motor, and any external forces such as friction or air resistance.

5. How can an elevator's acceleration be controlled?

An elevator's acceleration can be controlled by adjusting the motor's power and speed, as well as the weight limit of the elevator. Additionally, modern elevators use computer systems to precisely control acceleration and deceleration for a smoother ride.

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