Determining Friction of Table with 100g, 75cm, 2s

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Homework Help Overview

The discussion revolves around determining the friction of a table using a cart system influenced by a suspended mass. The problem involves concepts from dynamics, specifically relating to forces, acceleration, and frictional forces acting on the cart and the suspended mass.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the need to calculate acceleration using distance and time, and the relationship between applied forces and friction. There are attempts to clarify the roles of different forces acting on the masses involved, including tension and gravitational forces.

Discussion Status

Some participants have provided general and specific approaches to the problem, discussing the equations that relate the forces and accelerations. There is an ongoing exploration of the relationships between the normal force, gravitational force, and frictional force, with participants seeking clarification on these concepts.

Contextual Notes

Participants are navigating the complexities of the forces involved, including the assumptions about the pulley system and the need to account for both masses in the equations. There is a focus on understanding how to relate the forces to find the friction coefficient.

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1. A cart travels along a smooth table. The cart moves as a result of a suspended mass attached to a pulley that pulls the cart towards the end of the table. If the mass, displacement and time from start to finish are know, how can you determine the friction of the table?

I think you require acceleration for later on so use v=d/t and then a=v/t to find the acceleration. Then maybe use Fnet=ma or Fnet=Ff + Fapp, what is the applied force equal to and how would you determine it?

Numbers: 100g suspended weight. Distance of 75cm, total time of 2 seconds.
 
Last edited:
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Thoughts? It would help a lot :)
 
Draw a FBD.

I will give a general approach and a specific approach. In both these ignore rotational inertia of pulley

General approach: (better use this aid for solving)

There are masses M (on table) and m(dangling from pulley).

Consider equilibrium of each block. Find the forces acting.

For eg: the table block has inertial force M*a, frictional force F and string tension T

The mass m has inertial force m * a (note that the strings are assumed to be un extensible) and string tension T. Note that string tensions are the same as there is no friction to alater the tension.


Specific approach:
If you look closely, the gravitational force of mass m is pulling the mass M and m

So the applied force of mg is overcoming the inertial force Ma + ma + frictional force

(M + m)*a + F = m*g ---- (1)

a and F are unknowns

We also know the dispalcement of cart "S", time taken to reach the edge "t"

Use the equation S = (1/2)*a*t*t to find "a"
Put that "a" in (1) and find F

Now F = mu * Mg

mu = F/Mg
 
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Thanks, I'm pretty sure I understand. So the M (mass of cart) multiplied by a +force of friction= mass of suspended weight multiplied by gravity. After finding acceleration and plugging everything in you can find F which is the force of Friction, correct?
 
PhysicsAdvice said:
Thanks, I'm pretty sure I understand. So the M (mass of cart) multiplied by a +force of friction= mass of suspended weight multiplied by gravity. After finding acceleration and plugging everything in you can find F which is the force of Friction, correct?

Yes, but since both masses are accelerating at the same value, you must add both M and m.

so it is like this:

the ( M (mass of cart) + mass of block hanging ) multiplied by a +force of friction= mass of suspended weight multiplied by gravity
 
great, got it. Thanks !
 
Sorry to bother you again but is mg= to the normal force? as I know Fn=fg=mg and Ff=mu(Fn), but you said mg is the fapp, so in this case is Fn also Fapp?
 
PhysicsAdvice said:
Sorry to bother you again but is mg= to the normal force? as I know Fn=fg=mg and Ff=mu(Fn), but you said mg is the fapp, so in this case is Fn also Fapp?

Normal force Fn is the reaction force due to the mass M and "g". (weight of the cart)

The friction force is proportional to the Normal force, the constant of proportionality is the "mu". The applied force has to overcome BOTH the frictional force + inertial forces (M+m)*a
 

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