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You are not logged in. #1 20080611 19:00:45
Sliding blocksTriangle 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 20100222 20:20:05
Re: Sliding blocksAssuming 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). #3 20100222 23:12:49
Re: Sliding blocksIf θ_{1} and θ_{2} 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 (20100223 01:14:37) #4 20100222 23:18:39
Re: Sliding blocks
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 20100223 17:31:23
Re: Sliding blocksHaving looked this up on the net I agree that my suggested solution was way off base. 