Two stacked objects sliding down inclined plane.

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A 60kg boy rides down a 19º icy slope on a 2.6kg scale, with no friction between the scale and the hill. The discussion centers on calculating the static friction force exerted by the scale on the boy, which is determined to be zero due to the lack of relative motion between them. The boy and the scale slide down together, and the forces acting on the boy include gravity, inertia, normal force, and static friction. The participants clarify that the scale is not an inertial frame of reference, leading to the conclusion that static friction is indeed zero. The final consensus is that the initial misunderstanding about the interaction between the objects was incorrect.
michaelwiggin
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Homework Statement


"A 60kg boy rides down an icy hill of 19º slope while standing on a 2.6kg flat-bottomed bathroom scale. Assume there is no frictional force between the bottom of the scale and the hill. The static friction force the scale exerts on the boy is ____.


Homework Equations


... F = ma
Ff=\muFn
Not sure what else... really its the theory of it that's getting me.


The Attempt at a Solution


Well, my understanding is that if there was no friction between the boy and the top of the scale, then while the scale was sliding down the hill, the boy would be sliding downward not only with respect to the hill (as a result of the acceleration of the scale) but also with respect to the scale. But because it asks for the static friction of the scale on the boy, this must mean he's not moving, and thus when determining the friction we can use the usual Ff=mgsin\theta equation.


Thus 60kg(9.8m/s2)(sin19) = 191.4

Any mistakes that you all notice?

Thanks!
-Michael
 
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Both the boy and the scale slide downhill together. What is the acceleration of the boy? What are the forces acting on the boy?

ehild
 
>>Well, my understanding is that if there was no friction between the boy and the top of the scale
no, there is no frictional force between the scale and the ice hill

>>this must mean he's not moving
yes, relative the the scale he is not moving
 
The scale is not an inertial frame of reference as it is accelerating with respect to the stationary ground. If you stick to the scale as frame of reference, there is a force of inertia -ma acting on each body of mass m where a is the acceleration of the scale along the slope. You have gravity, the force of inertia, normal force and static friction acting on the boy. The sum of these forces must give zero.

ehild
 
You're correct ehild, my initial thinking was completely foolish. Somehow I believed that if frictionless plates (any objects I suppose) were stacked atop each other upon an inclined plane, than *each* accelerated with respect to the surface that they were placed on. That's absurd, of course, because if 100 plates were stacked then the top one would be accelerating at a rate of 981 m/s^2.*facepalm*The force of static friction is 0. Thanks fellas.
-Michael
 
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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