The Ex pert TAIH https:// usi38ok.theexpertt /Common/TakeTutorialAssignment.aspx

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The Ex pert TAIH https:// usi38ok.theexpertt /Common/TakeTutorialAssignment.aspx a.com (7%) Problem 13: Three glasses are placed by a waiter on a light-we ght tray. The first glass has a mass of Mi 650 g and is located R 15 cm from the center of the MI degrees above the positive x-axis. The second glass has a tray at an angle 45 mass of M 325 g and is located Rs 1 cm from the center of the tray at an angle 62 45 degrees below the positive x-axis. The third glass has a mass of M3 225 g and is located R 18 cm from the center of the tray at an angle 83 35 degree above the negative x-axis. A fourth glass of mass M 825 g is to be placed on the tray so that the center of mass is located at the center of the tray. Assignment Status Randomized Variables Click here for detailed view 650 15 cim Otheexpertta.com 45 degree Problem Status 325 Completed. 21 cim Completed 62 45 degrees Completed 225 Completed. 18 cim Partial 35 degree Completed 825 Completed 17% Part (a) Write a symbolic equation for the horizontal position from the central x-axis that the fourth glass must be placed so that the Completed horizontal center of mass of the four glasses is at the center of the tray Completed Grade Summary 10 Completed. M1 R1 (01) M2 R2 cos (02) M3 (03) Deductions 11 Completed. Potential 94% 12 Completed cos cos cos eo cosce1 7 8 9 HOME Sub 13 Partial Attempts remaining 14 per attempt 15 iled 2:39 AM 0/24/2015

Explanation / Answer

given that (x1,y1) = (0.15*cos(45) ,0.5*sin(45)) = (0.106,0.106)

(x2,y2) = (0.21*cos(45),-0.21*sin(45)) = (0.148,-0.148)

(x3,y3) = (-0.18*cos(35),0.18*sin(35)) = (-0.147,0.103)

and Xcm = 0

then Xcm= (m1*x1 + m2x2 + m3x3 = m4x4)/(m1+m2+m3+m4)

0 = (m1*x1 + m2x2 + m3x3 + m4x4)/(m1+m2+m3+m4)

0 = (m1*R1cos(theta1)) + (m2*R2*cos(theta2) - (m3*R3*cos(35) +( m4*x4)

x4 = -[(m1*R1cos(theta1)) + (m2*R2*cos(theta2) - (m3*R3*cos(35)] / m4

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