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#1 2008-06-11 19:00:45
- Kurre
- Power Member
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Sliding blocks
Triangle PQR with area A is given in a plane parallell to the gravitational acceleration g, with the length QR perpendicular to g, and the point P is above QR. Now we let two blocks (considered as points) slide down PQ and PR. This gives the different times t1 respective t2 for each block. If we assume there is no friction, find the minimum of T=t1+t2 and determine the triangle when this occurs.
#2 2010-02-22 20:20:05
- cberry
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Re: Sliding blocks
Assuming no friction, t1=t2 for any position of P (as they would fall at the same rate, due to gravety). Mimimum for t1+t2 then must be where PQ + PR are minimum, eg when P is midway along QR and where area A approaches the limit of 0. (cant be 0 as no longer a triangle).
-or-
Given: time = distance / acceleration.
We know
1 acceleration (due to gravety) is fixed
2 distance refers to vertical distance only (as no friction)
therefor Time is minimum when vertical distance is minimum. When point P approaches line of QR. And as that happens the shape of the triangle approaches the shape of the line QR
#3 2010-02-22 23:12:49
- JaneFairfax
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Re: Sliding blocks
If θ1 and θ2 are the angles at Q and R respectively and h is the , then
for i = 1, 2 (as can be deduced by application of elementary physics). As the area of the triangle is A, we have
You wanna minimize that?
Last edited by JaneFairfax (2010-02-23 01:14:37)
#4 2010-02-22 23:18:39
- JaneFairfax
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Re: Sliding blocks
cberry wrote:
Assuming no friction, t1=t2 for any position of P (as they would fall at the same rate, due to gravity).
That is not correct. 
The time taken for an object to fall a vertical distance h is
but if it slides down a frictionless slope inclined at angle θ to the horizontal, the time taken for it to fall the same vertical distance is
(as you can easily work out). So unless θ is 90°, it will take longer to slide down a slope than to fall freely under gravity.
#5 2010-02-23 17:31:23
- cberry
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Re: Sliding blocks
Having looked this up on the net I agree that my suggested solution was way off base.
'Jane', could you have a look at:
"Looking for a solution to a tricky little Hasselhoff problem..."
Posted 2010.02.20