Basic Forces Question from 9th grader - Hard to put in words

In summary, the problem asks for the tension force on the lower end of a rod with mass of 0.6 kg. The rod connects two blocks, the first / upper one has a mass of 5.0 kg while other/lower one has a mass of 4.0 kg. There is an upward force of 150 N applied on the whole system.The attempted solution is to use the mass of the rod only and find the tension force. The alternative solution is to use the mass of the combined assembly, which is 9.6 kg. The net force is 54 N and the acceleration is 15.625 m/s^2.
  • #1
Rod Alexei
29
0

Homework Statement


I have here a problem asking me to find the tension force on the lower end of a rod with mass of 0.6 kg. The rod connects two blocks, the first / upper one has a mass of 5.0 kg while other/lower one has a mass of 4.0 kg. There is an upward force of 150 N applied on the whole system.

Homework Equations


Given:
F = 150 N
Mass of whole system : 9.6 kg (mass of upper block+mass of rod + mass of lower)
a = 150 N/9.6 kg
a= 15.625 m/s^2

The Attempt at a Solution


Knowing that the system has acceleration:
Fnet= ma
Fnet = Tb (tension force on point b/ lower end of rod) - (Weight of rod + Weight of lower block)
Tb - (Wrod + Wlowrblck) = ma
HERE IS MY QUESTION (only capitalised for emphasis)
Tb - (5.88 N + 39.2 N) = ma
Tb - 45.08 N = m (15.625 m/s^2)
At this juncture I was having 2nd thoughts on what mass to use but since this problem is in multiple choice, I had to try out some things:
Tb - 45.08 = (4.6 kg)(15.625 m/s^2) -------> the mass I used is the sum of the masses of rod and lower block
Tb = 116.96 N -------> This answer was too large and was not part of the choices.
I had an alternative:
Tb - 45.08 = (0.6 kg)(15.625 m/s^2) ---------> this time I plugged in only the mass of the rod
Tb = 54.455 or approximately 54 N -------> This answer of mine was part of the choices so decided to stick with it.
If my 2nd solution is correct, why should I only consider the mass of the rod and not include the mass of the lower block?
Thank You very much
 
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  • #2
Welcome to Physics Forums.

It's very hard to picture this. Can you provide a diagram?

I assume the force is applied to the bottom of the stack? Have you drawn a free body diagram on the lower block and shown the forces acting on it?

Chet
 
  • #3
I would expect that since there is a reference to tension, vis-a vis compression, which to me implies a stretching of the connecting rod that the force is applied to the upper mass.
 
  • #4
OK. So, again, have you drawn a free body diagram on the lower mass and shown the forces acting on it? Based on this free body diagram, what is the force balance on the lower mass?

Chet
 
  • #5
Mod note: Copied diagram from Word doc and added it as an attachment.
Here is the diagram of the problem. Thank you for replying.
FBD.png
 

Attachments

  • Diagram of problem.docx
    15.2 KB · Views: 213
Last edited by a moderator:
  • #6
the answer is 62.5 N my friend
Who's going to calculate gravity as an external force(as the 150 N force acts upwards,gravity acts on the whole system downwards)
still make the question a bit clear,would you?
 
  • #7
My friend,it's obvious gravity's there
Please redo your question and i'll tell you something called the centre of mass method to solve such problems(:))
 
  • #8
UchihaClan13 said:
My friend,it's obvious gravity's there
Please redo your question and i'll tell you something called the centre of mass method to solve such problems(:))

It is not at all obvious that gravity is present; it was not mentioned in the original problem. The whole thing could be here on Earth in the presence of gravity, or it could be in outer space where there is no significant gravitational force on such a small object.

Certainly the OP should clarify this issue!
 
  • #9
Assuming, as UchichaClan13 suggests, that gravity is present.

You already determined that the mass of the combined assembly is 9.6 kg. Using the equation w = mg, what is the weight of the combined assembly?
From your free body diagram, there are two external forces acting on the assembly: 150 N and the weight (gravitational force) mg. What is the net force? What is the acceleration?

Next, you are going to have to do a free body diagram on just the lower block so that you can determine the force of the rod acting on this block. From Newton's 3rd law, how would this force compare with the tension force that the lower block applies to the rod?

Chet
 
  • #10
upload_2016-1-27_18-19-1.png
 
  • #11
Ray Vickson said:
space where there is no significant gravitational force on such a small object.
that's true but my friend,surely the question would have been elaborated further by the OP wouldn't it?
It's normally assumed that when we say that the 150 n force pulls the system upwards ,there's gravity acting downwards!
 
  • #12
UchihaClan13 said:
that's true but my friend,surely the question would have been elaborated further by the OP wouldn't it?
It's normally assumed that when we say that the 150 n force pulls the system upwards ,there's gravity acting downwards!

Normally, yes, but who knows if that is the case here? I refuse to assume; I would rather have the OP state clearly all the conditions of the problem. I know, I know... if one speaks of upward and downward that typically refers to the direction of gravity, but again, I am unwilling to make an assumption. Much better, I think, would be to force the OP to deal with the issue!
 
  • #13
Ray Vickson said:
Normally, yes, but who knows if that is the case here? I refuse to assume; I would rather have the OP state clearly all the conditions of the problem. I know, I know... if one speaks of upward and downward that typically refers to the direction of gravity, but again, I am unwilling to make an assumption. Much better, I think, would be to force the OP to deal with the issue!
Suppose that you add gravity to the free body diagram based on some unknown local gravitational field strength. Does that field strength change the final answer?
 
  • #14
jbriggs444 said:
Suppose that you add gravity to the free body diagram based on some unknown local gravitational field strength. Does that field strength change the final answer?
$$150-9.6g=9.6a\tag{overall force balance}$$
where a is the acceleration of the assembly. So,
$$a=\frac{150}{9.6}-g$$
$$T-4g=4a\tag{force balance on lower mass}$$
where T is the tension exerted by the rod on the lower mass, and the tension exerted by the lower mass on the rod.
So,
$$g+a=\frac{T}{4}=\frac{150}{9.6}$$
So, $$T=62.5 N$$

Yikes, you're right. The answer to the problem does not depend on whether gravity is included or not.
 
  • #15
Chestermiller said:
So,
T=62.5NT=62.5N​
T=62.5 N

Yikes, you're right. The answer to the problem does not depend on whether gravity is included or not.
See told you
gravity assumption would yield the correct answer
:)
 

Related to Basic Forces Question from 9th grader - Hard to put in words

1. What are the four basic forces?

The four basic forces are the gravitational force, electromagnetic force, strong nuclear force, and weak nuclear force.

2. How do these forces interact with matter?

The gravitational force acts on all matter with mass, the electromagnetic force acts on charged particles, the strong nuclear force binds protons and neutrons in the nucleus of an atom, and the weak nuclear force is responsible for radioactive decay.

3. Why are these forces considered "basic"?

These forces are considered "basic" because they are fundamental interactions that govern all physical processes in the universe.

4. How do scientists study and understand these forces?

Scientists study and understand these forces through experiments, mathematical equations, and theories such as Newton's laws of motion and Einstein's theory of relativity.

5. What are some real-world examples of these forces in action?

Examples of these forces in action include gravity holding objects to the Earth's surface, magnets sticking to metal objects, nuclear reactions in power plants, and radioactive decay in nuclear medicine.

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