Solving Newton's Problem: 2 Masses Stacked on Each Other

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The discussion focuses on a physics problem involving two stacked masses on a frictionless incline, where the goal is to determine the force required to keep the top block in equilibrium. Participants emphasize the importance of identifying the forces acting on the 10kg block, including its weight and the normal force from the incline. The conversation highlights the need to apply Newton's second law to analyze the acceleration of both the block and the incline. Confusion arises regarding the direction of acceleration and the components of the forces involved. Ultimately, understanding the relationship between the forces and the motion of both masses is crucial for solving the problem effectively.
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Homework Statement


Ok, so I know how to solve ALL the other types of Newton problems that have been thrown at us. I can't understand conceptually when two masses are stacked on each other..I've tried to google it relentlessly and am going into my prof's office hours.

The surfaces between the 10kg block and 20kg incline are frictionless.
What is the force that has to be exerted on the incline so the 10kg block does not move up or down (aka is in equilibrium). The force is directed so that the incline is pushed to the left.
The problem is attached...
I've tried making free-body diagrams but to no avail - I just don't understand what forces exist in the free body diagrams! Is the force applied to the incline in two components relative to the block? I'm just so confused...


Homework Equations



Newton's third law


The Attempt at a Solution


Eh...many pages of useless scribble..
 

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Here's a hint or two to get you going. Concentrate on the block. What forces act on it? See if you can figure out what acceleration it must have in order not to slide down the incline.
 
Forces acting on the block - Normal force, Weight in two components (mgsintheta, mgcostheta), Contact force between incline and block ?
 
Myronnie said:
Forces acting on the block - Normal force, Weight in two components (mgsintheta, mgcostheta), Contact force between incline and block ?
The contact force between incline and block is the normal force. So only two forces act on the block: its weight and the normal force from the incline.

Now apply Newton's 2nd law. Which direction is the block being accelerated?
 
the block is being accelerated to the right opposite the horizontal component of weight (mgsintheta)..but the incline is being accelerated to the left by the force? I'm still lost =.=
 
Myronnie said:
the block is being accelerated to the right opposite the horizontal component of weight (mgsintheta)..but the incline is being accelerated to the left by the force? I'm still lost =.=
If the force F is pushing both block and incline to the left, what direction are both accelerated?
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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