Dynamics Coursework; Velocity/Acceleration/Kinetic Friction

  • Thread starter DTskkaii
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  • #1
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Homework Statement


The 125kg concrete block A is released from rest in the position shown and pulls the 200kg log up the theta=30degrees ramp.

Case 1: if the coefficient of kinetic friction between the log and the ramp is Uk=0.5
(a) Find the magnitudes of velocity and acceleration of the block A as it hits the ground at B

Case 2: If it is changed to a log with 400kg, consider kinetic friction between the log and ramp Uk=0.5. Determine the value of the critical slope (theta), that if theta is larger than this value, the log will slip down when the concrete block A is released.

Case 3: If theta=30degrees and it is changed to a log of 400kg, determine the value of the friction coefficient (the critical friction coefficient) that would result in a zero acceleration of the block.


The Attempt at a Solution


Attachments:
'pic' shows a diagram of the problem
'001' shows my cable and FBD for how i have set up my equations
'002' shows my working so far for the first question

Attempts:
Case 1: I have achieved 2 equations as shown in 002
0.5(173.205)-2T+200sin30=(200/9.8)Ac
or 186.6025-2T=(200/9.8)Ac
and 125-T=(125/9.8)Aa

I currently have 2 equations with 3 unknowns, and dont quite know how to approach that.
For Case 1, I simply have to work out those 3 variables and then find velocity of Block A using Aa and fall distance, but really need a little assistance working out those set of equations.

Am currently working on other problems until I can get Case 1 done, then I will post attempts at Case 2 and 3, but if anyone can allude as to how I should approach them when I get there, that would be fantastic too.

Thankyou!
 

Answers and Replies

  • #2
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Replying because it doesnt seem that the attachments worked.
 

Attachments

  • pic.jpg
    pic.jpg
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  • 001.jpg
    001.jpg
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  • 002.jpg
    002.jpg
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  • #3
18
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I also found the third equation to use as 0=2Aa+2Ac.

Using this I found T, and substituted it in to find Aa and Ac.
However, by this method Aa and Ac and equal with opposite signs (as I should have predicted from the above formula). This is incorrect, so i'm stumped again.
 

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