What is the Acceleration of an Inclined Plane due to Recoiling Block?

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SUMMARY

The acceleration of an inclined plane due to a recoiling block is determined by analyzing the forces acting on both the block and the plane. The derived formula for the acceleration A of the plane is A = - (mg cos θ sin θ) / (M + m sin² θ), where m is the mass of the block, M is the mass of the inclined plane, and θ is the angle of inclination. The negative sign indicates the direction of acceleration relative to the chosen coordinate system. The discussion emphasizes the importance of clearly defining positive directions in physics problems to avoid confusion.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Familiarity with inclined plane mechanics
  • Knowledge of non-inertial reference frames
  • Basic trigonometry, particularly sine and cosine functions
NEXT STEPS
  • Study the effects of friction on inclined planes
  • Learn about non-inertial reference frames in classical mechanics
  • Explore vector components in physics problems
  • Investigate other dynamics problems involving recoiling objects
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Students studying physics, particularly those focusing on mechanics and dynamics, as well as educators looking for examples of inclined plane problems and their solutions.

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Homework Statement


A block of mass m is on an inclined plane of mass M, inclined at angle θ, and slides on the plane without friction. Find the acceleration of the plane.

The Attempt at a Solution


I am using the usual cartesian coordinate system with no rotations and letting up and forward be the positive directions. Let A be the acceleration of the plane as it accelerates backwards from the recoil due to the moving block. Define a non - inertial reference frame that is co - moving with the plane. The equations of motion for the block in this frame are m\ddot{y} = Ncos\theta - mg,m\ddot{x} = F_{apparent} = Nsin\theta - mA and we have, in this co - moving frame, the constraint \ddot{y} = -tan\theta \ddot{x}. The equation of motion for the plane in this frame is 0 = F'_{apparent} = -Nsin\theta - MA. Combining the equations for the block we get that N = mgcos\theta + mAsin\theta so MA = -Nsin\theta = -mgcos\theta sin\theta - mAsin^{2}\theta therefore A = -(\frac{mgcos\theta sin\theta }{M + msin^{2}\theta }). The book has the same answer except it is positive. I'm not sure why mine is negative. They don't really list if they are taking the backwards direction to be positive or not so I don't know if that is all there is to the issue. Thanks.
 
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Hi Isaac
There is nothing in the statement that would give you a preferred direction, I read it as finding the magnitude of the acceleration of the plane. Otherwise, since the acceleration is a vector in the end, you could as well wonder about which component is which.
So I don't think there is any issue at all, you solved the problem :)
 
Assuming you have exactly reproduced the statement of the problem from the book, I would guess you are supposed to take the positive direction as being whichever way the wedge moves. You say you let "up and forward be the positive directions", but you don't say forward for which object. If you meant forward for the block then you would expect a negative result for the wedge.
(Good job getting the right answer - easy to go wrong with a question like this.)
 

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