Answer All Part A Find TA, the tension in string A. Express your answer in newto

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Answer All

Part A Find TA, the tension in string A. Express your answer in newtons using three significant figures. Hints The figure (Figure 1) shows a model of a crane that may be mounted on a truck.A rigid uniform horizontal bar of mass m1 = 90.0 kg and length L = 5.70 m is supported by two vertical massless strings. String A is attached at a distance d=1.20 m from the left end of the bar and is connected to the top plate. String B is attached to the left end of the bar and is connected to the floorAn object of mass m2 = 3000 kg is supported by the crane at a distance T = 5.50 m from the left end of the bar. DVD AEas a 6 = ? Throughout this problem, positive torque is counterclockwise and use 9.80 m/s for the magnitude of the acceleration due to gravity. TA = 159417 Submit My Answers Give Up Incorrect; Try Again; 5 attempts remaining Figure 1 of 1 Part B Find TB, the magnitude of the tension in string B. Express your answer in newtons using three significant figures. m In Hints IVO AED tra O a ? TB = Submit My Answers Give Up

Explanation / Answer

net torque about B = 0

TA*d - m1*g*L/2 - m2*g*x = 0

TA*1.2 - (90*9.8*1.2/2) - (3000*9.8*5.5) = 0

Tension in string A = TA = 135191 N

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along vertical

In equilibrium net force = 0

TA - TB - m1*g - m2*g = 0

135191 - Tb - 90*9.8 - (3000*9.8) = 0

Tb = 104909 N <<<---------------ANSWER

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part A

Young's modulus Y = FL/(A*dL)

F = m*g

Y = mg*L/(A*dL)

dL = m*g*L/(A*Y)

A = area of cross section = pi*r^2

r = 2.5/2 = 1.25 cm = 0.0125 m

dL = 1500*9.8*800/(pi*0.0125^2*10^10)

dL = 2.39 m   <<<<------------ANSWER

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