Register to reply

Combined Mass

by Joran Verlaeck
Tags: combined, mass
Share this thread:
Joran Verlaeck
#1
Apr25-13, 03:09 PM
P: 2
Dear Forum Users,

I am studying physics on my own, i am new to this forum, i am working every example in the book and i have a question. I hope that you are so kind to give me an explanation so that i can continue my chapter exercises.

The question of the book is : What horizontal force must be applied to a large block of mass M shown in Figure P5.67 so that the tan blocks remain stationary relative to M ? Assume all surfaces and the pulley are frictionless. (Notice that the force exerted by the string accelerates M1)

See attachement for picture.

The calculation goes as follows.

M1*g - T = 0 =>
T = M2 * a =>
M1*g = M2 * a => a = (M1*g)/(M2)
F = TotalMass * a => in the book and on the internet the result is
F = (M + M1 + M2) * (M1*g) / (M2)

What i was wondering is if M1 is not connected by the string with M2, and the surface is frictionless, normally only block M would slide together with M1 and M2 would remain stationary and eventually fall off. The tension in the robe is provided by the gravitational force, no gravity
no tension. So in the end result F only needs to accelerate M and M1 and not M2 because that
is accelerated by gravity -> equation a = M1 * G / M2.

So why is the result (M + M1 + M2) * (M1*g) / (M2) and not (M + M1) * (M1*g) / (M2).
Because F is not responsible to give M2 that acceleration just for keeping M relative to M2 in rest.
So they both need the same acceleration.

Can you please help me with this reasoning.
Thx

Jöran
Attached Thumbnails
Picture.png  
Phys.Org News Partner Science news on Phys.org
Scientists develop 'electronic nose' for rapid detection of C. diff infection
Why plants in the office make us more productive
Tesla Motors dealing as states play factory poker
tiny-tim
#2
Apr25-13, 04:17 PM
Sci Advisor
HW Helper
Thanks
tiny-tim's Avatar
P: 26,148
Hi Jöran! Welcome to PF!
Quote Quote by Joran Verlaeck View Post
So why is the result (M + M1 + M2) * (M1*g) / (M2) and not (M + M1) * (M1*g) / (M2).
Because F is not responsible to give M2 that acceleration just for keeping M relative to M2 in rest.
So they both need the same acceleration.
I see what you mean.

But no, for two reasons:

i] suppose the three masses were fixed to each other …
the result would be the same, wouldn't it?
ii] this is an exercise in applying Newton's second law …

you must always apply it to all the external forces on a particular body (or bodies)

if you apply it to M and M1 combined, then you must include the external force (on M and M1) from the horizontal part of the rope …
(M + M1)a = F - T = F - M2a,

ie (M + M1 + M2)a = F
SammyS
#3
Apr25-13, 04:25 PM
Emeritus
Sci Advisor
HW Helper
PF Gold
P: 7,819
Quote Quote by Joran Verlaeck View Post
Dear Forum Users,

I am studying physics on my own, i am new to this forum, i am working every example in the book and i have a question. I hope that you are so kind to give me an explanation so that i can continue my chapter exercises.

The question of the book is : What horizontal force must be applied to a large block of mass M shown in Figure P5.67 so that the tan blocks remain stationary relative to M ? Assume all surfaces and the pulley are frictionless. (Notice that the force exerted by the string accelerates M1)

See attachement for picture.

The calculation goes as follows.

M1*g - T = 0 =>
T = M2 * a =>
M1*g = M2 * a => a = (M1*g)/(M2)
F = TotalMass * a => in the book and on the internet the result is
F = (M + M1 + M2) * (M1*g) / (M2)

What i was wondering is if M1 is not connected by the string with M2, and the surface is frictionless, normally only block M would slide together with M1 and M2 would remain stationary and eventually fall off. The tension in the robe is provided by the gravitational force, no gravity
no tension. So in the end result F only needs to accelerate M and M1 and not M2 because that
is accelerated by gravity -> equation a = M1 * G / M2.

So why is the result (M + M1 + M2) * (M1*g) / (M2) and not (M + M1) * (M1*g) / (M2).
Because F is not responsible to give M2 that acceleration just for keeping M relative to M2 in rest.
So they both need the same acceleration.

Can you please help me with this reasoning.
Thx

Jöran


Hello Joran Verlaeck. Welcome to PF !


If m1 and m2 are stationary relative M, then they all have the same acceleration, call it a .

So the net force, F, on the system must be such that F = (m1 + m2 + M)a . It's simply the application of Newton's 2ND Law of Motion.

haruspex
#4
Apr25-13, 07:45 PM
Homework
Sci Advisor
HW Helper
Thanks
P: 9,863
Combined Mass

Quote Quote by Joran Verlaeck View Post
in the end result F only needs to accelerate M and M1 and not M2 because that is accelerated by gravity -> equation a = M1 * G / M2.
F is not responsible to give M2 that acceleration just for keeping M relative to M2 in rest.
T = m1g = m2a. At the pulley, the string exerts a force T vertically and horizontally on M. The horizontal forces on M are therefore F, m1a from contact with m1 and m2a from the pulley.
Joran Verlaeck
#5
Apr26-13, 12:39 PM
P: 2
All of you guys,

Thanks for presenting me the reasoning why this is the case.
After i read the responses i released that i made a mistake by following my reasoning.
I was thinking of the same physics situation only this time m1 rests on the floor, the outcome is like tiny-tim suggested the same. If so M1 feel the normal force and in reality there is really no tension in the robe, but once the object starts moving, and we want to make sure that M2 has no velocity relative to M, the tension in the robe gets m2.a and if a is exactly m1*g/m2 the normal force on m1 is zero. In this case you can see that F must be responsible to the motion of M2 and all the equations remains the same.

Thanks for answering this topic.


Register to reply

Related Discussions
Pi and E combined Linear & Abstract Algebra 15
2 protons have less mass than combined Special & General Relativity 3
Can SR and GR be combined. Special & General Relativity 6
Can these be combined? Introductory Physics Homework 7
Combined mass acceleration Introductory Physics Homework 11