Determining the Acceleration (Two objects in contact)

AI Thread Summary
Two crates, weighing 75kg and 110kg, are in contact on a horizontal surface with a 620N force applied to the lighter crate. The total mass of the system is 185kg, leading to an acceleration of 3.35m/s² when accounting for friction. The force exerted by crate x on crate z is 251.25N, while crate z exerts a force of 368.5N on crate x. Reversing the crates results in the same force values due to the unchanged mass and applied force. It is crucial to consider friction when calculating the net forces affecting acceleration and interactions between the crates.
yandereni
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1. The problem statement, all variables and given/known data
2 crates, 75kg(crate x) and 110kg(crate z) are in contact and at rest on a horizontal surface. A 620N force is applied on crate x . If the coefficient of friction is 0.15, calculate:
a.) the acceleration of the system
b.)the force that each crate exerts on each other
c.) repeat when the crates reversed

2. Homework Equations

Msys = (massx)+(massz)
Force = mass(acceleration)

The Attempt at a Solution



a. Msys = 75kg + 110kg
= 185kg
620N = 185kg(a)
a = 3.35m/s2

b.
cratex= 75kg(3.35m/s2)
=251.25N
cratez= 110(3.35m/s2)
=368.5N

c.
cratez= 110(3.35m/s2)
=368.5N
cratex= 75kg(3.35m/s2)
=251.25Nthanks in advance!
 
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It may help to draw a force diagram to help yourself with this. For instance, with part A, you have to include friction as a part of what's going on with the system. So when you add up all the forces, you have a slightly different answer than what you have because the frictional force reduces the net force on the blocks in the direction of their displacement, and therefore reduces the acceleration. This is also what you have to keep in mind for parts B and C.
 
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