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y theexpertta.com Top 10 Easiest Med... Student Center Q Home- Quora ine D AE Buy Modalert onlin ut a p... Home Student: edayanov@gmail.com My Class Management 1 Help chapter 9 Begin Date: 7/23/2018 12:00:00 AM-Due Date: 8/5/2018 11:00:00 AM End Date: 8/5/2018 11:00:00 AM (20%) Problem 1: A uniform beam of length L and mass M has its lower end pivoted at P on the floor, making an angle 0 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. tatus for iew Otheexpertta.com atus rtial .25%Part(a) write an expression for the y component Py of the force exerted by the pivot on the beam. 25% Part (b) Write an expression for the tension T in the horizontal cable AB ompleted Grade-100 % Feedback Cerrect Student Final Submission Correct Answer Not available until end date Grade Sammary Completed 0% 0% Deduction for Final Subemission Dedactions for Incorrect Submissions, Hints and Feedback [7) Student Grade-100-0-0 = 100% Submission History Hints Feedback All Date times ore displayend in Easters Stnelard Time Red ssmission date times indicute lote work Time Answer Date 25% Part (e) write an e pression for the x-component Pr of the force esered by the pivot on the team in erms of r tons, if the mass of the beam is 26 kg. the length of the beam is n m and the - 25% Punt (d) what is the tension in the horizontal cable, in angle is 21Explanation / Answer
c) Px = T = 5/4 * Mg / tan (theta)
d) T = 5/4 * Mg / tan theta
= 5 * 26 * 9.8 / (4 tan 21)
= 829.721 N
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