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Homework Statement
1 1.00-kg glider on a horizontal air track is pulled by a string at an angle theta. The taut string runs over a pulley and is attached to a hanging object of mass 0.500 kg as in the below picture. (a) Show that the speed vx of the glider and the speed vy of the hanging object are related by vx = uvy, where u = z(z2 - h02)-2.
(b) The glider is released from rest. Show that at that instant the acceleration ax of the glider and the acceleration ay of the hanging object are related by ax = uay. (c) Find the tension in the string at the instant the glider is realeased for h0 = 80.0 cm and theta = 30.0 degrees.
Homework Equations
The usual laws of motion at an introductory level.
The Attempt at a Solution
Eh, I'm stymied. I've been so for a good year or so (like that other problem -__-). Here's my scratch work for (a) and (b). I found (c) fine enough.
T - 0.5g = 0.5ay -Tcos(theta) = ax = -T(z2 – h0^2)^.5/z -uax = T -uax - 0.5g = 0.5ay
Since nothing seemed to work, I decided to try to introduce a new force, "Ry" to represent the force the air pushed upwards on the glider. I also tried using the old kinematics equations. Note that "g(t)" and so forth is actually g * t--not a function of g at time t.
Ry +Tsin(theta) – g = 0 = Ry + Th0/z – g -T(z2-h02)1/2/z(t) = vx vy = 2T(t) – (Th0/z)(t) = 2T(t) - 0.5g(t) vy/(2T-(Th0/z)) = uvx/(T) = vy/(2 – h0/z) = -uvx vx = vy/(uho/2z – u) = 2zvy/(uh0 – 2zu)
(c) is easy enough. Tcos(30.0) = (1.00)ax = uay (0.500)g - T = (0.500)ay --> g - 2.00T = ay Now (in meters), z = h0/sin(30.0) = 1.6, so u = 1.6(1.62 - 0.82)-0.5 = 1.15 (approximately).
Thus, Tcos(30.0) = uay = 1.15(g - 2.00T) = 1.15g - 2.30T --> Tcos(30.0) + 2.30T = (cos(30.0) + 2.30)T = 1.15g. Solving for T then gives 3.56.
So, any pointers or hints for the first two parts? It was from an easy introductory course, so I'm probably missing something obvious.
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