lass Management Help mework 8 Begin Date (896) Problem 8: A uniform beam of leng
ID: 2038476 • Letter: L
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lass Management Help mework 8 Begin Date (896) Problem 8: A uniform beam of length Z- 1.6 m and mass M-17 kg has its lower end fixed to pivot at a point P on the floor, making an angle 0-24 as shown in the digram. A horizontal cable is attached at its upper end B to a point A on a wall. A box of the same mass Mas the beam is suspended from a rope that is attached to the beam one-fourth I from its upper end. : 3/29/201 8 12:00:00 AM-Due Date: 4/9/2018 í í:59:00 PM End Date: 4/9/2018 1?59:00 PM Ctheexpertta.c 33% Part (a) what is they-component P, of the force, in newtons, exerted by the pivot on the beam? Grade Summary Deductioas Potential10 sino l cosO | tano l (1.171819 cotan0 asacos0 tan0 acotan sinho cosh0 tan cotanh 4 5 6 Submissions Attempts remainin per attempt) etailed view Degrees O Radians Hint i give up Submit Feedback: deduction per feedback Hints: 2tt deduction per hint. Hints remaining 2 33% Part (b) Write an expression for the tension Tin the horizontal cable AB. 33% Part (c) what is the x-component P, of the force, in newtons, exerted by the pivot on the beam? i,Explanation / Answer
(A) balancing forces in Y- direction.
(system is in equilibrium hence Fnet = 0 and Net torque = 0 )
Fnet = Py - Mg - Mg = 0
Py = 2 M g = 2 x 17 x 9.8 = 333.2 N ......Ans
(B) balancing moment about point P,
= - (L/2)(Mg)(cos(theta)) - (3L/4)(Mg)(cos(theta)) + (L)(T)sin(theta) = 0
T sin24 = 5 M g cos24 / 4
T = (5 x 17 x 9.8 x cos24) / (4 sin24)
T = 467.7 N
(C) Net force in Horizontal = 0
Px - T = 0
Px = 467.7 N
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