Solving for Tension in Net Force Equation: Need Help!

In summary: I.In summary, in case I, if you know the direction of the acceleration, you should include it in the equation. If you don't know the direction of the acceleration, you can follow the same rule with the additional directive "keep it simple".
  • #1
Dun Artorias
1
0
Homework Statement
A wheel is mounted in a physics lab as shown above. The wheel has a mass M = 2.2 kg and radius R = 35 cm. The wheel is in the shape of a ring with rotational inertia I = MR^2. A object of mass m = 5 kg is attached to the end of a string which is wrapped around an inner hub (on the left hand side) with radius r = 2.7 cm. The object is released from rest a distance h = 77 cm above the floor and begins to move downward.

Determine the linear acceleration of the object.
Relevant Equations
Net Force = ma
Net Torque = I * alpha
I have tried finding the tension of the string through the equation
Net Force = T - mg = ma
but I am struggling with coming up with the correct acceleration. Is my net force equation correct?
 
Physics news on Phys.org
  • #2
:welcome:

Can you include a diagram for the problem?
 
  • #3
Dun Artorias said:
Is my net force equation correct?
It depends which way you are defining as positive for the acceleration.
What about a corresponding equation for the wheel?
 
  • #4
I'm finding that the hanging mass will experience a constant linear acceleration less than ##g## If I've interpreted it correctly.
 
  • #5
I agree with @haruspex that your equation is ambiguous but not because it depends on whether you define the acceleration positive when it is up or positive when it is down. It's because, apparently, you didn't incorporate the known direction of the acceleration in your equation. See the explanation below which is lengthier than originally planned and I apologize for that.

Here we have a 1 D vector equation expressing Newton's second law. Using arrows over symbols to indicate vectors, $$\sum_i \vec{F}_i=m\vec a.$$Now $$\vec T+m\vec g=m\vec a$$Note that a "plus" sign appears between all vectors because the net force is the sum of all forces in Newton's law. In 1D, the equation is simplified by removing the arrows to convert the vector equation into a scalar equation. There are two possibilities.

Case I: You know the directions of all vectors in the 1D equation.
First you pick a direction as positive. Second, you replace the arrows with "+" sign in front of the vector symbol if it points along the chosen positive direction and a "-" sign if the vector points opposite to the chosen positive direction. In this particular problem you know that the direction of the tension is "up", the direction of the weight is "down" and the direction of the acceleration is also "down." Following this rule,
  1. If you choose "up" as positive ##~\vec T+m\vec g=m\vec a \rightarrow T-mg=-ma.##
  2. If you choose "down" as positive ##~\vec T+m\vec g=m\vec a \rightarrow -T+mg=ma.##
Note that one equation becomes the other if you multiply both sides by ##-1## which is equivalent to choosing the opposite direction as positive. Also, symbols ##T##, ##g## and ##a## in the equations stand for magnitudes of vectors and are positive quantities. If you replace symbols with given numbers in either equation and the acceleration comes out negative, you will know that you made a mistake somewhere. The magnitude of a vector is positive independently of the choice of axes.

Case II: You don't know the direction of one vector in the 1D equation.
You follow the same rule with the additional directive "keep it simple". We can modify the situation in this problem a bit and say the we have a vertically accelerating mass ##m## attached to a string in which the tension is ##T##. We are asked to find the acceleration, a vector, in terms of the other quantities.

First note that the tension is "up" and the weight are "down", therefore the tension and the weight must appear with a relative negative sign in the expression for the net force regardless of whether "up" is positive or negative. Newton's second law can be written in vector equation form (a bit unconventionally but clearly as to what is meant) $$\frac{1}{m}T\text{(up)} +g\text{(down)}=a\text{(?)}$$ We can remove the directions that we know and replace them with negative signs. For the acceleration, we "keep it simple" and assume that the unknown direction denoted by the question mark is positive as defined by the choice of positive. Note that here ##a## stands for the scalar component of the vector not its magnitude; it could be positive or it could be negative, we just don't know. We get
  1. ##a=\dfrac{1}{m}T -g~## if "up" is positive
  2. ##a=-\dfrac{1}{m}T +g~## if "down" is positive
Either equation is correct and equation 1 is the form that you have posted. Note that one does not get one equation from the other if both sides are multiplied by ##-1##, i.e. they are not the same equation. That is, I think, what @haruspex meant when he said
haruspex said:
It depends which way you are defining as positive for the acceleration.
Nevertheless, if you substitute numbers in either equation and the acceleration turns out to be a negative number, it will mean that the initial assumption (when the question mark was removed) was incorrect. The acceleration must be opposite to the chosen positive direction. The catch here is that if you make a calculation error and the acceleration comes out with one sign when it should be the other, you wouldn't know it. That is why if you know the direction of the acceleration, you should put it in as shown in case I.
 
  • #6
Since @Dun Artorias wrote T-mg, it is clear that up is being taken as positive for T. The remaining question is whether up is also positive for the acceleration.
kuruman said:
Case I: You know the directions of all vectors in the 1D equation.
First you pick a direction as positive.
  1. If you choose "up" as positive ##~\vec T+m\vec g=m\vec a \rightarrow T-mg=-ma.##
  2. If you choose "down" as positive ##~\vec T+m\vec g=m\vec a \rightarrow -T+mg=ma.##
I don't understand that. If |T|>mg and up is chosen as positive (1) will give a negative a, meaning it accelerates down?
Guessing correctly which way things will move is of no benefit. The equations are the same.

With up positive for everything, T-mg=ma.

With down positive for everything, it depends whether you think of T as the magnitude of the tension in the string section or as the scalar value of the force exerted on the mass by that tension.
In the magnitude view, -T+mg=ma; in the scalar view, T+mg=ma, with T being negative.
 
  • #7
Dun Artorias said:
Homework Statement:: A wheel is mounted in a physics lab as shown above.
:oldconfused::oldconfused::oldconfused:
 
  • Like
Likes erobz
  • #8
Lnewqban said:
:oldconfused::oldconfused::oldconfused:
I wouldn’t hold your breath, you’re turning blue.
 
  • #9
erobz said:
I wouldn’t hold your breath, you’re turning blue.
I see a vectorial discussion developing on one unique post and a promised picture that never made it into this thread.
Please, see post #2.
 
  • #10
Lnewqban said:
I see a vectorial discussion developing on one unique post and a promised picture that never made it into this thread.
Please, see post #2.
I was trying to make a joke about the color of your emojis…I’m afraid it didn’t land.
 
  • #11
erobz said:
I was trying to make a joke about the color of your emojis…I’m afraid it didn’t land.
Oh no, please; I got the joke, and I liked it. :smile:
My comment was not on that.

Perhaps others can, but I can't visualize the problem's set up only from the OP's description.
A posted diagram, as requested in post #2, would be nice.
 
  • Like
Likes erobz
  • #12
Lnewqban said:
Oh no, please; I got the joke, and I liked it. :smile:
My comment was not on that.

Perhaps others can, but I can't visualize the problem's set up only from the OP's description.
A posted diagram would be nice.

I picture wheel fixed to the wall on an axle, and a string wrapped around a small hub on the axle with the weight attached to it dropping to the floor.

A posted diagram would be nice, but the OP is MIA.
 
  • Like
Likes Lnewqban
  • #13
I believe that I have finally seen it:

image.jpg
 
  • Like
Likes erobz and kuruman

FAQ: Solving for Tension in Net Force Equation: Need Help!

1. What is the basic formula for solving tension in a net force equation?

The basic formula for solving tension in a net force equation is derived from Newton's Second Law, F = ma. For a system with multiple forces, you sum the forces acting on the object. The tension (T) in the rope or string can be found by setting up the equation T - mg = ma for vertical motion or T = ma for horizontal motion, where m is the mass, g is the acceleration due to gravity, and a is the acceleration of the system.

2. How do you account for multiple objects connected by a string or rope?

When dealing with multiple objects connected by a string or rope, you need to consider the forces acting on each object separately and then use the constraint that the tension is the same throughout the string. For example, if two objects of masses m1 and m2 are connected by a string over a pulley, you would write separate equations for each mass and solve them simultaneously, ensuring that the acceleration of both masses is the same.

3. What role does friction play in solving for tension?

Friction plays a significant role in solving for tension, especially in scenarios involving surfaces. If friction is present, you need to include the frictional force in your net force equation. For example, if an object is being pulled on a surface with friction, the equation would be T - f_friction = ma, where f_friction is the frictional force, which can be calculated as f_friction = μN, with μ being the coefficient of friction and N the normal force.

4. How do you solve for tension in a system with an inclined plane?

In a system with an inclined plane, you need to resolve the forces parallel and perpendicular to the plane. The tension can be found by considering the components of gravitational force along the plane. The net force equation along the plane would be T - mgsin(θ) = ma, where θ is the angle of the incline. You also need to consider the normal force and friction if applicable.

5. Can tension be negative, and what does it signify?

Tension cannot be negative. If your calculations result in a negative value for tension, it usually indicates that you've chosen the wrong direction for the forces in your initial setup or that the assumed direction of acceleration is incorrect. Tension is a pulling force, so it should always be positive or zero.

Back
Top