The Velocity of Block B Relative to A

  • Thread starter kiwifruit
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In summary, After applying force P, block B accelerates to the right at 3m/s^2. When B has velocity of 2m/s, the velocity of B relative to A is 3/4 of the velocity of A. The acceleration of B relative to A is 3m/s^2. The absolute velocity of point C is unknown, as it depends on the values of Sb and Sa. The total length, L, is equal to 3 times the difference between Sb and Sa, plus the difference between Sd and Sa. Differentiating this equation with respect to time yields 0 = 3 times the derivative of Sb minus Sa with respect to time, plus the derivative of Sd
  • #1
kiwifruit
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After applying force P, block B accelerates to the right at 3m/s^2
When B has velocity of 2m/s determine:
1.The velocity of B relative to A
2.The acceleration of B relative to A
3.The absolute velocity of point C

The image attached show how this works.
I worked out the total length,L to get

L = 3(Sb - Sa) + (Sd - Sa)
so if you differentiate with respect to t you get

0 = 3 d(Sb - Sa)/dt + d(Sd - Sa)/dt

then I am stuck.

or is the differential equation supposed to be
L = 3(Sb - Sd) + 4(Sd - Sa)
so differentiating would get
0 = 3 d(Sb - Sd)/dt + 4 d(Sd - Sa)/dt

because somehow Va=3/4Vb
if its the second differential equation could someone explain how to get from there to get Va and Vb?
 

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  • #2
hi kiwifruit! :smile:

(try using the X2 icon just above the Reply box :wink:)
kiwifruit said:
L = 3(Sb - Sa) + (Sd - Sa)
so if you differentiate with respect to t you get

0 = …

you're making this very complicated :redface:

Sd is constant, so use the equation for L to find Sa as a function of Sb :smile:
 

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