Dynamics: Time taken for Block A to slide on Block B

  • Thread starter jdawg
  • Start date
  • #1
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Homework Statement


Find the time it takes for block A to slide 0.5m on block B.
Mass A=10 kg
Mass B=35 kg
I included a screen shot from my online homework.

Homework Equations


FR=FNμk


The Attempt at a Solution


For Block A:
mAa=T-FRA-WAsin(30)
T=10a+68.67

For Block B:
FRC=(mAg+mBg)μBC
FRA=mAAB
mBa=T+FRC-FRA-WAsin(30)-WBsin(30)

After substituting in all my values:
35a=T-196.2
T=35a+196.2

Setting my Tensions equal to each other:
10a+68.67=35a+196.2

magnitude of a=5.1012

Finding time:
s-s0=0.5at2+v0t
s0=0
v0t=0

0.5=.5(5.1012)t2

t=0.443s

Which is wrong... I can't figure out what exactly I'm doing wrong.
 

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Answers and Replies

  • #2
Doc Al
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For Block B:
FRC=(mAg+mBg)μBC
FRA=mAgμAB
mBa=T+FRC-FRA-WAsin(30)-WBsin(30)
(1) Careful with signs: What are the directions of those frictional forces on block B?
(2) Careful with signs: What is the direction of the acceleration of block B? (Since you use "a" as the magnitude of the acceleration, apply the signs correctly.)
(3) Why does the weight of A enter into an equation for B?
 
  • #3
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2
Ok, I see what you mean with the friction forces.
Does the weight of block A not put a force on block B?
 
  • #4
Doc Al
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Does the weight of block A not put a force on block B?
Certainly block A exerts both a normal force and a friction force on block B. But the weight of block A is a force that acts on block A, not B.
 
  • #5
367
2
Ok! So my new equation for B is this:
mBa=T+mBAB+mBBC-mBgsin(30)

Plugging in values...
T=35a+68.67

I still think I messed up, when I set these equations equal to each other, my acceleration ends up being 0 :(
 
  • #6
Doc Al
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45,089
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Ok! So my new equation for B is this:
mBa=T+mBgμAB+mBgμBC-mBgsin(30)
Note my previous item (2): Take care with the sign of the acceleration. There is a constraint: If block A moves up the incline, how must block B move?
 
  • #7
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I agree that block B should move down the incline. I'm not sure which part of my equation you're saying needs the sign change.
 
  • #8
Doc Al
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I agree that block B should move down the incline. I'm not sure which part of my equation you're saying needs the sign change.
Block A has acceleration of magnitude "a" going up the incline; Block B has the same magnitude of acceleration going down the incline. Adjust your signs accordingly. (Compare the direction of acceleration with the directions of the forces.)
 

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