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Consider these two cases:

1)An object of mass $m_1$ is on a movable cart of mass $m_2$, coefficient of friction between the object and cart is known.If a force $F$ acts on the object and it's not strong enough to overcome the friction, then the acceleration of the cart and the whole system is $a=\frac{F}{m_1+m_2}$.
However, if the force is strong enough to overcome friction, the acceleration of the cart is actually equal to $\frac{F_{friction}}{m_2}$.How does friction affect the cart and where is mass of the object ($m_1$)?I don't remember any textbook telling this about friction forces.

2)Similar example but external force is now gravitational force.This time the cart is sloped.The force acting on the sloped cart isn't just due to friction between the cart and object, but due to the resultant force acting on the object that is on the slope (sum of gravitational force and friction force).Why is second (gravitational) force considered in this example but not in the previous one (first example considers only the friction but not the force $F$).

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The force on the object in second example is $F=m_2gsin(\alpha)-\mu m_2gcos(\alpha)$
And the horizontal force is
$F_x=Fcos(\alpha)$
The force that pushes the cart (in second example) is equal to $F_x$ but has opposite direction

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In the first case, if static friction has been overcome, then kinetic friction acts forward on the cart and backward on the object, which must be considered separately. In the second case there is a normal force from the object acting on the cart. This is caused by a component of gravity in that direction acting on the object. This normal force has a horizontal component acting to the right. The kinetic friction force acting on the cart has a horizontal component acting to the left. In the absence of other external forces, the resultant of these two accelerates the cart to the right. In the static case, the resultant would be zero. Note that if the object is sliding down the incline, its direction of motion is not parallel to the surface of the moving cart, which must be used as the basis for an accelerating frame of reference.

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  • $\begingroup$ I think I understand some of this but I was surprised that friction force acts on other objects as well.I'm confused after the part "This normal force has a horizontal component acting to the right. The kinetic ... has component acting to the left.Are you sure you haven't swapped directions for these?I tried to find horizontal components of these forces but they dont match the one below the picture.For the horizontal gravitational component, I got $F_{gx}=-sin(\alpha)cos(\alpha)mg$, but for horizontal friction I got $F_{rx} = \mu mg$, as can be seen here - i.imgur.com/stVnVZN.png $\endgroup$ – user3711671 Sep 5 '19 at 18:59
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    $\begingroup$ Friction involves action and reaction. $\endgroup$ – R.W. Bird Sep 5 '19 at 19:41
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Case 1- How does friction affect the cart and where is mass of the object ($m_1$)?

It is the friction force that $m_1$ exerts on $m_2$ that is responsible for the horizontal acceleration of $m_2$ as there are no horizontal forces acting on $m_2$ other than the friction force between $m_1$ and $m_2$, assuming there is no friction between the cart and the ground. You can see this on the free body diagram below for $m_2$. What's more, the mass of the object $m_1$ is included in the calculation of the friction force that mass $m_1$ exerts on $m_2$. The following additional explanation and the diagram below is offered:

In order for $m_1$ to slip on $m_2$, the applied force $F$ must exceed the maximum static friction force, or

$$F>μ_{s}m_{1}g$$

where $μ_{s}$ is the coefficient of static friction and $m_{1}g$ is the downward force of $m_1$ on $m_2$ due to gravity which is equal and opposite to the upward force exerted by $m_2$ on $m_1$.

Once relative motion occurs it is kinetic (sliding) friction that opposes the relative motion and is equal to

$$F_{friction}=μ_{k}m_{1}g$$

where $μ_{k}$ is the coefficient of kinetic (sliding) friction. In general, $μ_{k}<μ_{s}$.

The kinetic friction force is acting backwards on $m_1$ and forwards on $m_2$.

From the free body diagram below for $m_2$ we see that the only force acting on it, assuming no friction between the cart and the ground, is the friction force that $m_1$ exerts on $m_2$. Therefore the acceleration $a_2$ of $m_2$ is

$$a_{2}=\frac {F_{friction}}{m_{2}}=\frac{μ_{k}m_{1}g}{m_2}$$

So you should see from the free body diagram on $m_2$ below that it is only the friction force that is responsible for accelerating $m_2$, and mass $m_1$ is taken into consideration because its weight bearing down on $m_2$ is the source of the friction force on $m_2$.

Case 1- Why is second (gravitational) force considered in this example but not in the previous one (first example considers only the friction but not the force $F$).

The answer to the previous question should answer this. As already discussed above, gravity was considered in case 1 because it contributes to the friction force on $m_2$. You just need to substitute $F_{friction}=μm_{1}g$ where $g$ is the acceleration due to gravity. But the downward force of gravity in the first case has nothing to do with the force $F$ applied horizontally to $m_1$. In the second case, the force $F$ as well as the friction force, are functions of the gravitational force.

Hope this helps.

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