How Do You Solve a Friction and Motion Problem Involving a Mug and a Tablecloth?

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In summary, the student has attempted to solve a homework problem involving friction and acceleration of an object, but made a mistake.
  • #1
Zynoakib
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Homework Statement


Q69
20150722_173223.jpg


Homework Equations


F=ma
v = u + at
v^2 = u^2 + 2as
s = ut + (1/2)at^2

The Attempt at a Solution


Find the deceleration by friction:
F = ma
0.1 = (0.2)a
a = 0.5 ms^-2

and the acceleration of the mug would be 3 - 0.5 = 2.5 ms^-2

Find the time required for the tablecloth to travel 30 cm
u = 0, , s = 0.3, a = 3, t = ?
s = ut + (1/2)at^2
0.3 = (1/2)(3)(t^2)
t = 0.447 s

Find the displacement of mug within 0.447s
t = 0.447, s = ?, u = 0, a = 2.5
s = ut + (1/2)at^2
s = (1/2)(2.5)(0.2)
s= 0.25 m , but the answer is 0.06 m

Thanks!
 
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  • #2
Zynoakib said:
Find the deceleration by friction:
F = ma
0.1 = (0.2)a
a = 0.5 ms^-2

and the acceleration of the mug would be 3 - 0.5 = 2.5 ms^-2
In which direction is the mug accelerating? In which direction does friction between mug and tablecloth act? And what, exactly, do you mean by "a" here?
 
  • #3
Zynoakib said:
Find the displacement of mug within 0.447s
t = 0.447, s = ?, u = 0, a = 2.5
s = ut + (1/2)at^2
s = (1/2)(2.5)(0.2)
s= 0.25 m , but the answer is 0.06 m

Thanks!

Is the acceleration of block 2.5m/s-2 in ground frame ?

Hope this helps .
 
  • #4
Qwertywerty said:
Is the acceleration of block 2.5m/s∧-2 in ground frame ?
If you draw a free body diagram for the block, what forces act on it?
 
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  • #5
Zynoakib said:
and the acceleration of the mug would be 3 - 0.5 = 2.5 ms^-2

2.5m/s-2 is acceleration of block relative to cloth .
 
  • #6
Qwertywerty said:
2.5m/s-2 is acceleration of block relative to cloth .
Good. And its acceleration relative to the ground is?
 
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  • #7
jbriggs444 said:
Good. And its acceleration relative to the ground is?

0.5m/s-2 - Thanks jbriggs444 .
 
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  • #8
why is the acceleration of the mug relative to the ground not (2.5 + 3) because the acceleration of the tablecloth is 3 and the mug is sitting on it and accelerate with 2.5 ms^2, so the acceleration of the mug relative to the ground should be 5.5 ms^2

but even if I use a = 0.5 to calculate, I still cannot get the answer. Why's that?
 
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  • #9
Okay - you seem to be confused .

Try answering these questions :

jbriggs444 said:
If you draw a free body diagram for the block, what forces act on it?

then

Qwertywerty said:
Acceleration of block relative to cloth is ?

and then

jbriggs444 said:
And its acceleration relative to the ground is?
 
  • #10
Zynoakib said:
but even if I use a = 0.5 to calculate, I still cannot get the answer. Why's that?

Use a = 0.5m/s2 only in your last step - you will get your answer .

Zynoakib said:
Find the displacement of mug within 0.447s
t = 0.447, s = ?, u = 0, a = 2.5
s = ut + (1/2)at^2
s = (1/2)(2.5)(0.2)
s= 0.25 m , but the answer is 0.06 m

Thanks!
 
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  • #11
Thanks, I have got the answer!

But if I want to make the calculation process easier (without the free body diagram), can I just think it in this way?
Since the friction between the mug and the tablecloth is 0.1N, meaning that the force transferred from the tablecloth to the mug for the mug to move can only be 0.1N, so I should

F = ma
0.1 = 0.2a
a = 0.5 ms^-2
 
  • #12
Zynoakib said:
But if I want to make the calculation process easier (without the free body diagram), can I just think it in this way?
Since the friction between the mug and the tablecloth is 0.1N, meaning that the force transferred from the tablecloth to the mug for the mug to move can only be 0.1N, so I should

F = ma
0.1 = 0.2a
a = 0.5 ms^-2

Force is not transferred , it is exerted . And no , you don't need the FBD for this question - it was only for you to visualize a bit better .

The rest is fine .
 
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  • #13
Your working is wrong.
The correct solution is:
mass of mug = 200g = 0.2kg
mug located = 30cm = 0.3m (from the edge of the table)
Acceleration of cloth = 3m/s^2
Using s = ut + 1/2at^2
Finding time:
0.3 = 0+1/2*3*t^2
t = 0.2 sec

Finding the how far mug can move.
s = ut + 1/2at^2
s = 0 + 1/2x3x(0.2)^2
s = 0.06m

The answer is 0.06m.
 
  • #14
Hercules said:
Using s = ut + 1/2at^2
Finding time:
0.3 = 0+1/2*3*t^2
t = 0.2 sec

You've made a mistake here - t2 = 0.2 sec .

Also , try to not solve the entire question - You can read about this in the guidelines .
 
  • #15
Hercules said:
Your working is wrong.
The correct solution is:
mass of mug = 200g = 0.2kg
mug located = 30cm = 0.3m (from the edge of the table)
Acceleration of cloth = 3m/s^2
Using s = ut + 1/2at^2
Finding time:
0.3 = 0+1/2*3*t^2
t = 0.2 sec

Finding the how far mug can move.
s = ut + 1/2at^2
s = 0 + 1/2x3x(0.2)^2
s = 0.06m

The answer is 0.06m.
That's completely wrong.
First, the given distance is not to the edge of the table, it's to the edge of cloth.
Second, your calculation of the time would find how long it takes for the cloth to move .3m. This does not tell you where the mug is in relation to the cloth at that time.
Third, you then used the acceleration of the cloth as though it were the acceleration of the mug. The only reason you did not end up with an answer of 0.3m is your fourth error, confusing t with t2, as Qwertywerty noted.

Fwiw, the quickest way is to note that the acceleration of the mug is 1/6 that of the cloth. Since t2 is the same for both, it travels 1/6 of the distance. The difference between the two distances is 0.3m.
 

FAQ: How Do You Solve a Friction and Motion Problem Involving a Mug and a Tablecloth?

What is the Law of Motion Problem?

The Law of Motion Problem is a concept in physics that describes the relationship between an object's motion and the forces acting upon it. It is based on Sir Isaac Newton's three laws of motion, which state that an object will remain at rest or in uniform motion unless acted upon by an external force, the force applied to an object is equal to its mass multiplied by its acceleration, and for every action, there is an equal and opposite reaction.

How do you solve a Law of Motion Problem?

To solve a Law of Motion Problem, you will need to apply the three laws of motion to the given situation. First, identify the forces acting on the object and their direction. Then, use Newton's second law to calculate the net force on the object. Finally, use this net force and the object's mass to calculate its acceleration. You can then use this acceleration to determine the object's motion.

What are some common examples of Law of Motion Problems?

Some common examples of Law of Motion Problems include a car accelerating or braking, an object being thrown or kicked, a person walking or running, and a roller coaster going up and down a hill. These situations can be analyzed using Newton's three laws of motion to understand the forces and motion involved.

What is the difference between a static and dynamic Law of Motion Problem?

A static Law of Motion Problem involves an object at rest or in uniform motion, while a dynamic Law of Motion Problem involves an object experiencing a change in its motion. In a static problem, the net force on the object is zero, while in a dynamic problem, the net force is not zero and results in a change in the object's motion.

What are some tips for solving Law of Motion Problems?

Some tips for solving Law of Motion Problems include drawing a free-body diagram to visualize the forces acting on the object, breaking the problem down into smaller parts, using proper units and significant figures, and double-checking your calculations. It is also helpful to have a good understanding of Newton's laws of motion and how to apply them to different situations.

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