A Block Slides Down an Inclined Plane

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A block sliding down an inclined plane with friction moves at a constant speed, indicating that the net force acting on it is zero. The forces at play include the weight of the block, the normal force perpendicular to the incline, and friction opposing the motion. The normal force acts at a 90-degree angle to the inclined surface, while friction acts parallel to the incline. The discussion concludes that, according to Newton's third law, the force exerted by the block on the inclined plane points straight down. This conclusion aligns with the understanding that the forces must balance to maintain constant velocity.
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Homework Statement


A block slides down an inclined plane with friction. It is sliding at a constant speed. The only forces acting on the block are its weight and contact forces due to the incline. The inclined plane is stationary.

The inclined plane is a wedge that is highest on the left (eg. you would slide down towards the right). That is the way the block is sliding.

Which way does the force exerted by the block on the inclined plane point?


Homework Equations





The Attempt at a Solution



Well, I think there's a friction force pointing along the inclined plane towards the upper left and a normal force 90 degrees to the inclined plane pointing down. Adding these gives one force, pointing towards the left (straight to the left).

But apparently the answer is straight down (like along the y axis). I don't understand why, can anyone help?
 
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jumbogala said:
Well, I think there's a friction force pointing along the inclined plane towards the upper left and a normal force 90 degrees to the inclined plane pointing down.
First consider forces acting on the block. Which way does friction and the normal force act on the block? What's the net force on the block?
 
The normal force acting on the block would be 90 degrees to the surface of the inclined plane, in a northwest direction.

The friction acting on the block would be opposite the block's motion, so it would be in a northeast direction parallel to the inclined plane.

Wouldn't the forces acting on the inclined plane, from the block, be the same as these but in the opposite direction?
 
jumbogala said:
The normal force acting on the block would be 90 degrees to the surface of the inclined plane, in a northwest direction.

The friction acting on the block would be opposite the block's motion, so it would be in a northeast direction parallel to the inclined plane.
I'm not sure how you are defining east and west, but OK. (Is the block sliding southeast or southwest?)

Wouldn't the forces acting on the inclined plane, from the block, be the same as these but in the opposite direction?
Absolutely.
 
I am defining east and west so that the block is sliding southeast.

Ok, so now if I add up two vectors, one in a southeast direction (but still 90 degrees to the surface of the inclined plane). And another in southwest direction (but still parallel to the inclined plane), the vector points straight down.

Which is the answer!

Thank you!
 
Excellent.

Another way to see it is this: We know that the net force on the block is zero, since it moves with constant velocity. The only forces on it are the force of the inclined plane and the force of gravity. Since gravity acts straight down, the inclined plane force must act straight up. Therefore, from Newton's 3rd law, the force of the block on the inclined plane must act straight down.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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