(10%) Problem 6: A uniform beam of length L and mass M has its lower end pivoted
ID: 1783614 • Letter: #
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(10%) Problem 6: A uniform beam of length L and mass M has its lower end pivoted at P on the floor, making an angle with the floor. A horizontal cable is attached at its upper end B to a point A on a wall. A box of mass M is suspended from a rope that is attached to the beam one-fourth L from its upper end ©theexpertta.com > 25% Part (a) Write an expression for the y-component Py of the force exerted by the pivot on the beam Grade Summary 0% 100% Submissions Attempts remaining: S cotan() sin( | sin() tan( % per attempt) detailed view Submit I give up! Hints: 1 % deduction per hint. Hints remaining: 3 Feedback: 1% deduction per feedback. 25% Part (b) Write an expression for the tension T in the horizontal cable AB 25% Part (c) Write an expression for the x-component Pr of the force exerted by the pivot on the beam, in terms of T 25% Part (d) What is the tension in the horizontal cable, in newtons, if the mass of the beam is 37 kg, the length of the beam is 13 m, and the angle is 49Explanation / Answer
part(a)
In equilibrium along vertical Fnet = 0
Py - Mg - Mg = 0
Py = 2Mg
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part (b)
In equilibrium net torque = 0
T*L*sintheta - Mg*L/2*costheta - Mg*3L/4*costheta = 0
T*sintheta = Mgcostheta*(1/2 + 3/4)
T*sintheta = Mg*costheta*(5/4)
T = 1.25*Mg/tantheta
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part (c)
along horizontal
Fnet = 0
Px - T = 0
Px = T = 1.25*Mg/tantheta
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part (d)
T = 1.25*37*9.8/tan49 = 394 N <<<<------ANSWER
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