Understanding Sign Conventions in Elevator Tension Problems

[Abu]

Homework Statement

An elevator and its load have a combined mass of 800 kg. The elevator is initially moving downward at 10.0 m/s; it slows to a stop with constant acceleration in a distance of 25.0 m. What is the tension T in the supporting cable while the elevator is being brought to rest

Relevant Equations

V^2 = Vo^2 +2a(X-Xo)
F = ma

So the first thing that I did was solve for the acceleration of the elevator, given the velocities and distances given in the question. My question regarding this issue is not so much the procedure needed to solve the question, but rather the sign conventions for the values given and calculated:

I'll say that up is positive and down is negative -

Given:

Vo = -10 m/s
V = 0 m/s
X = -25 meters
m = 800kg
Tension = ?

Solution:

Solving for acceleration
V^2 = Vo^2 +2a(X-Xo)
0^2 = (-10)^2 + 2a(-25)
0 = 100 - 50a
-100 = -50a
a = 2.00 m/s^2 <--- This value makes sense to me, because the elevator is accelerating upwards (positive direction) to slow down

Solving for tension:

F = ma
Ftension - mg = ma
Ftension = ma + mg
Ftension = 800(2) + 800(-9.8)
Ftension = -6240 N

I picked -9.8 for gravity because acceleration due to gravity is acting downwards, and I designated downwards to be negative

So this answer is incorrect for a few reasons:

1. Tension should be greater than the weight of the elevator since the elevator is accelerating upwards.
2. I said that upwards is positive, and since tension is acting upwards, it should be positive (not the negative value I got)

Can someone explain how come my sign conventions are incorrect? I'm not entirely sure where the flaw in my reasoning is.

Thanks so much for your time, patience, and consideration!

 

[hutchphd]

Homework Statement: An elevator and its load have a combined mass of 800 kg. The elevator is initially moving downward at 10.0 m/s; it slows to a stop with constant acceleration in a distance of 25.0 m. What is the tension T in the supporting cable while the elevator is being brought to rest
Homework Equations: V^2 = Vo^2 +2a(X-Xo)
F = ma

So the first thing that I did was solve for the acceleration of the elevator, given the velocities and distances given in the question. My question regarding this issue is not so much the procedure needed to solve the question, but rather the sign conventions for the values given and calculated:

I'll say that up is positive and down is negative -

Given:

Vo = -10 m/s
V = 0 m/s
X = -25 meters
m = 800kg
Tension = ?

Solution:

Solving for acceleration
V^2 = Vo^2 +2a(X-Xo)
0^2 = (-10)^2 + 2a(-25)
0 = 100 - 50a
-100 = -50a
a = 2.00 m/s^2 <--- This value makes sense to me, because the elevator is accelerating upwards (positive direction) to slow down

Solving for tension:

F = ma
Ftension - mg = ma
Ftension = ma + mg
Ftension = 800(2) + 800(-9.8)
Ftension = -6240 N

I picked -9.8 for gravity because acceleration due to gravity is acting downwards, and I designated downwards to be negative

So this answer is incorrect for a few reasons:

1. Tension should be greater than the weight of the elevator since the elevator is accelerating upwards.
2. I said that upwards is positive, and since tension is acting upwards, it should be positive (not the negative value I got)

Can someone explain how come my sign conventions are incorrect? I'm not entirely sure where the flaw in my reasoning is.

Thanks so much for your time, patience, and consideration!

You should draw a free body diagram for the elevator. Gravity acts down but the Tension acts up. on the elevator.

 

[Abu]

You should draw a free body diagram for the elevator. Gravity acts down but the Tension acts up. on the elevator.

Thanks for your response! I recognize that tension acts up, as that is actually how I realized my answer was incorrect (because tension is acting upwards, and upwards is positive, I should have gotten a positive value instead of my negative one). Gravity acts downwards, I agree, and that's also why I used -9.8 in my final calculation to calculate the tension. Unfortunately, I still do not know how come my signs are wrong.

 

[TSny]

F = ma
Ftension - mg = ma

This is excellent as long as you take the quantity mg to be a positive quantity. The minus sign in front of mg takes care of the direction of the force. Then, mg is the magnitude of the force which should be a positive number. So, g here should be considered the magnitude of the acceleration due to gravity (a positive number). Most textbooks that I've seen define g to be a positive number representing just the magnitude of the acceleration due to gravity. If you write -mg and also take g to be -9.8 m/s², then you've done a "double negative" making -mg a positive quantity. But that would represent an upward force, not a downward force.

 

[Abu]

This is excellent as long as you take the quantity mg to be a positive quantity. The minus sign in front of mg takes care of the direction of the force. Then, mg is the magnitude of the force which should be a positive number. So, g here should be considered the magnitude of the acceleration due to gravity (a positive number). Most textbooks that I've seen define g to be a positive number representing just the magnitude of the acceleration due to gravity. If you write -mg and also take g to be -9.8 m/s², then you've done a "double negative" making -mg a positive quantity. But that would represent an upward force, not a downward force.

Oh I see now! I completely forgot that net force would be the sum, and so what I really calculated was Ftension + (-mg). That makes sense, thanks so much. My textbook defined g to be a positive number as well, but that lead me to another question that is perhaps semi-related to this one...

If I were dropping an object from a certain height, and was using the kinematic equations to solve for a certain variable, would I still use positive g in my kinematic equations even though the object is being sent downwards? My velocities and displacements would be negative as usual since the object is falling, but what about gravity?

Thanks so much for your response, really appreciate it!

 

[TSny]

My textbook defined g to be a positive number as well, but that lead me to another question that is perhaps semi-related to this one...

If I were dropping an object from a certain height, and was using the kinematic equations to solve for a certain variable, would I still use positive g in my kinematic equations even though the object is being sent downwards?

Well, I would certainly take g to be positive since I always use the same convention as your textbook, where g denotes the magnitude of the acceleration due to gravity. Now, that might mean that in some situations I would need to write a negative sign in front of g in order to get a correct expression. For example, suppose I take my y-axis as pointing upward and I'm dealing with a ball thrown upward so that it rises and then falls. I might need to use the basic kinematical equation

v_y² = v_y0² + 2a_yΔy

For a_y, I would take a_y = -g for both the rising and the falling of the ball. (And also at the instant at the top of the flight!) When I substitute for g, I would substitute 9.8 m/s², never -9.8 m/s². This would yield a_y =-9.8 m/s². The negative value indicates that the acceleration is in the opposite direction of what I chose to be the positive direction. So, the acceleration is downward.

If for the same problem I chose the y-axis as pointing downward, then the only change would be to write a_y = g instead of a_y = -g. But, still, g itself would be a positive number. Now I get a_y =+9.8 m/s² indicating that the acceleration is in the same direction as I chose for the positive y-direction. So, again the acceleration is downward as it should be for gravity.

 

[Abu]

Well, I would certainly take g to be positive since I always use the same convention as your textbook, where g denotes the magnitude of the acceleration due to gravity. Now, that might mean that in some situations I would need to write a negative sign in front of g in order to get a correct expression. For example, suppose I take my y-axis as pointing upward and I'm dealing with a ball thrown upward so that it rises and then falls. I might need to use the basic kinematical equation

v_y² = v_y0² + 2a_yΔy

For a_y, I would take a_y = -g for both the rising and the falling of the ball. (And also at the instant at the top of the flight!) When I substitute for g, I would substitute 9.8 m/s², never -9.8 m/s². This would yield a_y =-9.8 m/s². The negative value indicates that the acceleration is in the opposite direction of what I chose to be the positive direction. So, the acceleration is downward.

If for the same problem I chose the y-axis as pointing downward, then the only change would be to write a_y = g instead of a_y = -g. But, still, g itself would be a positive number. Now I get a_y =+9.8 m/s² indicating that the acceleration is in the same direction as I chose for the positive y-direction. So, again the acceleration is downward as it should be for gravity.

Thanks so much for your help, I'll use that method for my future calculations and make a post if I run into issues again. Really appreciate your time!