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

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The discussion centers on calculating the acceleration of a system with two crates on a horizontal surface, where a 620N force is applied to a 75kg crate. The user confirms their calculation of acceleration as 1.9 m/s², which matches the book's answer, but they seek clarity on differing methodologies. Both the user's and the book's approaches are validated, emphasizing that the acceleration of the center of mass is determined by the net external forces divided by the total mass. The conversation concludes that understanding the forces at play, including friction, is crucial for accurate calculations. The methods discussed are both correct and yield the same result.
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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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