Calculating Acceleration of a System with Two Crates on a Horizontal Surface

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Homework Help Overview

The discussion revolves around calculating the acceleration of a system consisting of two crates with different masses on a horizontal surface, subjected to an external force and friction. The original poster explores different approaches to derive the acceleration, comparing their method with that presented in their textbook.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster outlines their method for calculating acceleration using net forces and friction, questioning the validity of their approach compared to the textbook method. Participants discuss the correctness of both methods and the underlying principles of force and acceleration.

Discussion Status

Participants generally agree that both methods presented are valid. There is an emphasis on understanding the relationship between applied forces, friction, and the resulting acceleration of the system. Some participants highlight the concept of center of mass in relation to the problem.

Contextual Notes

The original poster expresses uncertainty about their calculations and seeks confirmation of their logic, indicating a desire for deeper understanding rather than simply obtaining the correct answer.

Genericcoder
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I have a question on block that I was solving in my book.
I got the answer right but different procedures for doing it I want to make sure my logic is correct,and I didn't get the answer out of probability.



Two crates of mass 75kg and 110kg, are in contact and at rest on a horizontal surface. A 620N force is exerted on the 75kg crate. If the coefficient of kinetic friction is 0.15, calculate a)The acceleration of the system.



Here I solved as follows:-

Fnet1 = F1 - F12 - Ff1 --> F1 - f12 - Ff1 = (m1)(a);

Fnet2 = F21 - Ff2 --> F21 - Ff2 = (m2)(a);

We know since objects are in horizontal plane then normal force is equal to weight of objects.
Fn1 = (m1)(a); --> Ff1 = (u)(m1)(a);
Fn2 = (m2)(a); --> Ff2 = (u)(m2)(a);


Fnet1 -> F1 - F12 - (u)(m1)(a) = (m1)(a);
Fnet2 -> F21 - (u)(m2)(a) = (m2)(a);

Adding both equations together we'll get net acceleration on the system.

F1 - F12 - um1g + F21 - um2g = m1a + m2a;

F - (u)(m1)(g) - (u)(m2)(g) = (a)(m1 + m2);
(F - (u)(m1)(g) - (u)(m2)(g)) / (m1 + m2) = a;

Substituting in for values we'll get the following:-

a = 1.9 m/s^2 ,which is the same answer as in my book but my book got it as follows:-

Fx = F1 - Fr = (m1 + m2)(a);

F1 - Fr = (m1 + m2)(a) -> a = (F1 - Fr) / (m1 + m2);
Substituting for values he got that answer which is the same as mine.
So did I do something wrong here or maybe somewhere in my math could be modified to get that same formula.


Thanks.
 
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Your method is correct. The book might mean the force of friction on Fr and uses the result of a derivation, the same what you applied: The acceleration of the CM of a system of interacting bodies equals the resultant of the external forces divided by the total mass.

The external forces are the applied force F and the forces of friction (and the normal force and gravity which cancel each other).

ehild
 
Both methods look correct. Whichever way you view it, the applied force must overcome friction, and then the excess accelerates the combined bodies.
 
We are dealing with center of mass. So total mass is m1+m2.
The center of mass is accelerating.
 
Thanks guys that makes sense :).
 

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