How Do You Calculate the Velocity of Slider A at θ = 90°?

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Discussion Overview

The discussion revolves around calculating the velocity of slider A when it reaches an angle of θ = 90° in a mechanical system involving two sliders and a horizontal force. Participants are exploring the application of work-energy principles and kinematic equations to solve the problem, which is presented as a homework question.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant presents an initial attempt at a solution using work-energy principles, calculating work done and relating it to kinetic energy.
  • Another participant questions the distance used in the work done calculation and suggests incorporating gravitational potential energy (mgh) into the equations.
  • There is a discussion about the correct distance the force acts upon, with one participant asserting that the distance is only 0.2 m instead of 0.4 m.
  • Participants express uncertainty about the velocity of slider B and its impact on the calculations for slider A's velocity.
  • One participant recalculates the velocity of slider A based on updated assumptions but remains unsure about the accuracy of their result.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the correct approach to the problem, with multiple competing views on the distances involved and the contributions of different forces to the final velocity calculation.

Contextual Notes

There are unresolved assumptions regarding the distances traveled by the sliders and the application of forces, which affect the calculations presented. The discussion reflects varying interpretations of the problem setup and the equations involved.

dietwater
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Homework Statement



Each of the sliders A and B has a mass of 2 kg and moves with negligible friction in its
respective guide, with y being in the vertical direction (see Figure 3). A 20 N horizontal force
is applied to the midpoint of the connecting link of negligible mass, and the assembly is
released from rest with θ = 0°. Determine the velocity vA with which slider A strikes the
horizontal guide when θ = 90°.
[vA = 3.44 m/s]

http://http://img.photobucket.com/albums/v242/021003/tutorial12.jpg


Homework Equations



1/2 mv^2

F = ma

Wp = mgh

SUVAT


The Attempt at a Solution



When at 0 degrees

W=0J

At 90

F=20N

W = 20xd = 8J

Work from cart A = 0.5mv^2

Therefore 16 = mv^2

v = 2 rt2

Or...

do i need to add the energy from 20n force and from cart b...

0.5mv^2 (b) + 8J = 0.5mv^2 (A)

with F = ma, 20/10 a = 10 therefore v (b) = 2 rt 2

sub this into above eq.

8 + 8 = 0.5mv^2

v = 4

Help!

Iv been goin round in circles, clearly I am wrong lol can someone explain how i could work this out please
 
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Hi dietwater! :smile:

hmm … your d in your work done is wrong.

And what happened to mgh?
 
tiny-tim said:
Hi dietwater! :smile:

hmm … your d in your work done is wrong.

And what happened to mgh?

lol il start again with mgh added in! lol

Fd + mgh + o.5mv^2 (b) = 0.5mv^2 (a)

I don't know how to get d, I just assumed its 0.4 as its the distance it has to travel?
 
dietwater said:
I don't know how to get d, I just assumed its 0.4 as its the distance it has to travel?

No … the point of application of the force is only moving 0.2 :wink:
 
tiny-tim said:
No … the point of application of the force is only moving 0.2 :wink:

am i right in thinking v(b) is still 2rt2?

If so..

Fd + mgh + o.5mv^2 (b) = 0.5mv^2 (a)

20x0.2 + 2x9.81x0.4 + 0.5x2x(2rt2)^2 = 0.5x2x(v^2)

so v = 3.98

Im not sure where Iv gone wrong
 
Hi dietwater! :smile:

(have a square-root: √ and try using the X2 tag just above the Reply box :wink:)

(and your picture didn't show up because you put too many http//s in it :rolleyes:)
dietwater said:
am i right in thinking v(b) is still 2rt2?

Nooo :redface: …

imagine it's a cross-roads instead of a T-junction …

which way would B be going as A sails across? :wink:​
 

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