Two identical point masses, each of mass 'M' are connected to one another by a massless string of length 'L'. A constant force 'F' is applied at the mid-point of the string. If 'l' be the instantaneous distance between the two masses, what will be the acceleration of each mass?
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Assume that the masses are on a horizontal frictionless table. No gravity related forces. Let the instantaneous distance between the two masses = AB = d = L Sinθ.F = 2 M a_y => a_y = F/(2M)F = 2 T cosθ = net force on the system T = F/(2 cosθ)a_x = T Sinθ /M = F*Tanθ/(2M)
secθ = L/√(L² - d²)
Net acceleration = a = √[(a_x)²+(a_y)²]
a = F Secθ/(2M) =F L /[2 M √(L² - d²) ]
Assume that the masses are on a horizontal frictionless table. No gravity related forces. Let the instantaneous distance between the two masses = AB = d = L Sinθ.F = 2 M a_y => a_y = F/(2M)F = 2 T cosθ = net force on the system T = F/(2 cosθ)a_x = T Sinθ /M = F*Tanθ/(2M)
secθ = L/√(L² - d²)
Net acceleration = a = √[(a_x)²+(a_y)²]
a = F Secθ/(2M) =F L /[2 M √(L² - d²) ]
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kvnmurty:
done. i hope that is clear and easy
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