Calculating Tension in Friction & Pulley: Solving for a=3 m/s^2

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Homework Help Overview

The discussion revolves around calculating the tension in a string connected to a block on a flat car that is accelerating. The problem involves concepts from dynamics, specifically Newton's laws, friction, and the relationship between the acceleration of the block and the car.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the forces acting on the block, including tension and friction, and question how these forces relate to the block's acceleration. There are discussions about the direction of forces and the implications of the car's acceleration on the block's motion.

Discussion Status

Participants are actively engaging with the problem, raising questions about the relationships between forces and the motion of the block. Some have offered insights into the forces at play, while others are seeking clarification on specific aspects of the problem.

Contextual Notes

There is a noted lack of information regarding the mass of the flat car, which some participants express as a concern. Additionally, there are discussions about the assumptions made regarding the massless pulley and the nature of friction in this context.

Calculus Master
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1. If flat car is given an acceleration a = 3 m/s^2 starting from rest, compute tension (in N) in the light inextensible string connected to block A of mass 30 kg. Coefficient of friction between block and flat car is = 0.50.

Neglect mass of pulley and its friction. Take g = 10.
2. Friction = ## \mu N ##
F = ma
ma = mg - T
3. No idea. I couldn't approach the problem. Please tell me how to approach these types.
 
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Calculus Master said:
1. If flat car is given an acceleration a = 3 m/s^2 starting from rest, compute tension (in N) in the light inextensible string connected to block A of mass 30 kg. Coefficient of friction between block and flat car is = 0.50.

Neglect mass of pulley and its friction. Take g = 10.
2. Friction = ## \mu N ##
F = ma
ma = mg - T
3. No idea. I couldn't approach the problem. Please tell me how to approach these types.

Write the Newton's laws in the frame of the ground. It is an inertial frame. Write down all the forces(along both axes). Also observe that for every meter the car moves, block A moves by 2 meter.
 
Calculus Master said:
Take g = 10.
I hope the original problem gave some units for g!
 
In case pcm's post is not clear enough, it's the forces on the block that matter. You don't need to analyse forces on the cart since its motion is a given.
 
I also want it's answer.Please guide how to start?
 
gracy said:
I also want it's answer.Please guide how to start?
What are the horizontal forces acting on the block? What is the acceleration of the block? What equation does that give you?
 
haruspex said:
What are the horizontal forces acting on the block?
Friction and tension.
 
What is the acceleration of the block?
Same as acceleration of the car(truck)?
 
haruspex said:
What equation does that give you?
Net force on the block=mass of the block multiplied by acceleration of the truck.
 
  • #10
if the acceleration of the block = 2 * truck acceleration, then the force required in the string = m * a
then add the friction force
 
  • #11
dean barry said:
then add the friction force
why should I add?
 
  • #12
acceleration force is forward (say + ve) and friction force is rearward (say - ve), the tension force is the difference.
adding gets the same result.
 
  • #13
dean barry said:
acceleration force is forward (say + ve) and friction force is rearward (say - ve), the tension force is the difference.
adding gets the same result.
Yes, it produces the same equation, but I feel it would be less confusing for gracy to stick to the normal formulation ##\Sigma F_x = m a_x##.
gracy, do you see why the acceleration of the block is not the same as the acceleration of the truck?
 
  • #14
dean barry said:
acceleration force is forward (say + ve) and friction force is rearward
But according to
The page number 6 of attached pdf
the direction of friction force should be forward.
 

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  • #15
gracy said:
But according to
The page number 5 of attached pdf
the direction of friction force should be forward.
That's a different situation. In the textbook example, the only force accelerating the block is the friction, so of course it must be acting in the forward direction. In this thread, there is a string pulling on the block.
Friction always opposes relative motion of the two surfaces in contact. Which way does the block move relative to the truck?
 
  • #16
haruspex said:
Which way does the block move relative to the truck?
rearward.
 
  • #17
dean barry said:
if the acceleration of the block = 2 * truck acceleration
Why?
 
  • #18
gracy said:
rearward.
Yes, sorry, I meant in this thread, which way does it move relative to the "flat car".
 
  • #19
gracy said:
Why?
What is the acceleration of the block relative to the car?
Does the total length of the string change?
 
  • #20
haruspex said:
which way does it move relative to the "flat car".
After Considering pull on block by string?
 
  • #21
haruspex said:
Does the total length of the string change?
No,it is for sure.
 
  • #22
gracy said:
After Considering pull on block by string?
Of course.
 
  • #23
Tension in string is pulling the block in forward direction, right?
 
  • #24
gracy said:
Tension in string is pulling the block in forward direction, right?
Right.
 
  • #25
So,to find in which direction the block will move,I need to know force is larger in which direction,i.e tension +force by truck in forward direction should be greater than force of friction then only block will move in forward direction.
 
  • #26
gracy said:
So,to find in which direction the block will move,I need to know force is larger in which direction,i.e tension +force by truck in forward direction should be greater than force of friction then only block will move in forward direction.
You know which way the block will move relative to the car (since the string neither stretches nor breaks).
And what do you mean by "force by truck in forward direction"? There are only two horizontal forces on the block.
 
  • #27
I only know force of friction (static in nature) i.e 0.5 multiplied by 30 multiplied by 10=150 N.
 
  • #28
haruspex said:
what do you mean by "force by truck in forward direction"? There are only two horizontal forces on the block.
As truck is moving and block is in contact with the truck will not it exert any force?
 
  • #29
gracy said:
As truck is moving and block is in contact with the truck will not it exert any force?
Yes, friction. What else?
 
  • #30
When we board in bus ,we are initially (before getting in bus)at rest,but then start traveling at the speed of bus,As per Newton's first law some external must acts on us.What's that?Is it fiction ?
You may find this trivial,but seriously that's why I was thinking that flat car does exert force on block.
 

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