Acceleration of block on acc' wedge

  • #1
A 45 deg wedge is pushed along a table with constant acceleration A. A block of mass m slides without friction on the wedge find its acceleration. Clue if A=3g then [tex] \ddot{y} = g [/tex]
I'm not sure of the forces on the block, if A is parallel to the base of the wedge then what component of A acts on the block and what is its magnitude? I don't know. The force on the wedge is (M+m)*A, where M is the mass of the wedge. Noting that if A is large enough the block will stay put on the wedge as the wedge is accelerated away. The table is taken as the x-axis. [tex] \sum{F_x} = m*a_x [/tex]. Please see attached diagram in pdf format. Thanks in advance for the help.
 

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Answers and Replies

  • #2
ideasrule
Homework Helper
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Do you know how to calculate the acceleration of a block sliding down a frictionless ramp under normal gravity, without the ramp being accelerated?

If so, consider what happens when the ramp is accelerating. To the block, the acceleration feels exactly the same as a gravitational field of strength "a". You can pretend that a force of ma is being exerted on the block and calculate its component parallel to the ramp, just like you would do with the real gravity. Then, just add or subject this component to the force contributed by gravity, and you'll have the total force on the block.
 
  • #3
327
2
I do not agree that the force on the wedge is (M+m)*A. It would if the acceleration of the block was also A. But it's not.

I couldn't find anything wrong with your free body diagram of the block.

The following is the way I solved the problem.

I used your free body diagram of the block to apply Newton's second law in the x (horizontal) direction (eqn 1) and y (vertical) direction (eqn 2) to the block. The acceleration of the block in the x direction was taken to be [itex]\ddot{x}[/itex] and the acceleration of the block in the y direction was taken to be [itex]\ddot{y}[/itex].

I used Newton's third law to find the reaction force on the wedge due to contact with the block. Then I found the force on the block [itex]F[/itex] by applying Newton's second law horizontally (eqn 3).

Now I have three equations. I used equations 2 and 3 to find [itex]F[/itex] in terms of [itex]\ddot{y}[/itex] by eliminating the reaction force [itex]N[/itex]. Then I used the clue given in the problem statement with this to find a general expression for [itex]F[/itex]. So I had an expression for [itex]\ddot{y}[/itex].

I combined equations 1 and 2 by eliminating [itex]N[/itex] to find [itex]\ddot{x}[/itex] in terms of [itex]\ddot{y}[/itex].

Now I have [itex]\ddot{x}[/itex] and [itex]\ddot{y}[/itex], so I know the acceleration of the block.
 
  • #4
327
2
Oops, sorry ideasrule. My answer took so long to type that I didn't know you had given an answer first.
 
  • #5
So ideasrule: m*a = m*gsin(theta) if the wedge is stationary and a is the acceleration parallel to the ramp. Then for accelerating wedge [tex] F_{parallel} = mg\sin\theta + ma\cos\theta \mbox{ where } F_{parallel} \mbox{ is parallel to the ramp} [/tex]
[tex] F_{parallel} = F_{total} = \frac{mg}{\sqrt{2}} + \frac{ma}{\sqrt{2}} \mbox{ where } \theta = \frac{\pi}{4} [/tex]. Having the total force on the block I now need the force parallel to the table and the force perpendicular to it F_parallel = F_total*cos(theta) = m(g+a)/2
F_perpendicular = f_parallel*sin(theta) = -m(g+a)/2. Now I need to find a interms of A
 

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