If the mass of the moon were tripled and it was moved to half as far f

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Homework Help Overview

The discussion revolves around two distinct physics problems: one concerning the gravitational force between the Earth and the moon under altered conditions, and the other involving the acceleration of a box on an inclined ramp with friction. The first problem is related to gravitational forces and celestial mechanics, while the second pertains to dynamics and frictional forces.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • The original poster attempts to confirm the calculation of gravitational force based on changes to the moon's mass and distance. They also seek verification of the acceleration of a box on a ramp, referencing forces acting on the box.

Discussion Status

Some participants have acknowledged the correctness of the gravitational force calculation. However, there is an ongoing discussion regarding the correct approach to finding the acceleration of the box, with some clarification needed on the components of forces acting on the box.

Contextual Notes

Participants are navigating assumptions about the forces involved in the second problem, particularly the application of trigonometric functions to resolve weight components. There is also a mention of the conditions under which static friction is not sufficient to prevent movement.

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>>>If the mass of the moon were tripled and it was moved to half as far from the centre of the Earth as it now rests, what would the new gravitational force between the Earth and the moon be?


Can someone confirm with me that the answer is 12 times the original?


>> A 12 kg box is sitting on a ramp which is angled at 30deg
to the horizontal. If the coefficient of kinetic friction is 0.20, what is the acceleration of the box down the ramp? (Assume static friction is not high enough to prevent it from moving.)

I got an answer of 20.36N, can someone double check that for me too?

fk = uk Fn = (0.2)(101.8N) = 20.36N

Fn = (12)(9.8)cos 30 = 101.8N
 
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Hi, oneshot. Welcome to PF.

Correct on the first question.

You still need to find the acceleration for the second question.
 
oh right, thanks
Fnet is mg cos 30 - fk right
and then use Fnet = ma and solve for a
 
oneshot said:
oh right, thanks
Fnet is mg cos 30 - fk right
and then use Fnet = ma and solve for a

Cos 30 is not correct for getting the component of the weight parallel to the ramp.
 

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