(33%) Problem 2: Three glasses are placed by a waiter on a light-weight tray. Th
ID: 1584134 • Letter: #
Question
(33%) Problem 2: Three glasses are placed by a waiter on a light-weight tray. The first glass has a mass of Mi 525 g and is located R1 17 cm from the center of the tray at an angle ,-45 degrees above the positive x-axis. The second glass has a mass of M-425 g and is located R2-24 cm from the center of the tray at an angle = 45 degrees below the positive x-axis. The third glass has a mass of M,-325 g and is located R,-19 cm from the center of the tray at an angle ,-45 degrees above the negative x-axis. A fourth glass of mass M 875 g is to be placed on the tray so that the center of mass is located at the center of the tray 3: Otheexpertta.com 8 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 horizontal center of mass of the four glasses is at the center of the tray. 17% Part (b) Calculate the numeric value of the horizontal position from the central x-axis of the fourth glass in cm. Grade Summary 0% 100% sin) tan( ) | cos) cotanasin) acos0 atan)acotan sinh0 tanh0 Submissions Atterupts remaining: 5 (0% per attempt) detailed view 2 3 cosh0 Degrees O Radians Submit Hint Hints: 0% deduction per hut. Hints remaining: Feedback: deduction per feedback. - 17% Part(c) write a symbolic equation for the vertical position from the central y-axis that the fourth glass must be placed so that the vertical center of mass of the four glasses is at the center of the tray. 17% Part (d) Calculate the numeric value of the vertical position from the central y-axis of the fourth glass in cm. 17% Part (e) Calculate the distance that the fourth glass is away from the center of the tray in cm. 17% Part (I) Calculate the angular position of the fourth glass relative to negative x-axis in degrees.Explanation / Answer
from the figure,
x1 = R1*cos(theta1) = 17*cos(45) = 12.02 cm
y1 = R1*sin(theta1) = 17*sin(45) = 12.02 cm
x2 = R2*cos(theta2) = 24*cos(45) = 16.97 cm
y2 = -R2*sin(theta2) = -24*sin(45) = -16.97 cm
x3 = -R3*cos(theta2) = -19*cos(45) = -13.4 cm
y3 = R3*sin(theta2) = 19*sin(45) = 13.4 cm
let x4, y4 are the coardinats of 4th mass.
a)
we know,
Xcm = (M1*x1 + M2*x2 + M3*x3 + M4*x4)/(M1 + M2 + M3 + M4)
0 = (M1*x1 + M2*x2 + M3*x3 + M4*x4)/(M1 + M2 + M3 + M4)
0 = (M1*x1 + M2*x2 + M3*x3 + M4*x4)
x4 = -(M1*x1 + M2*x2 + M3*x3)/M4
= -(M1*R1*cos(theta1) + M2*R2*cos(theta2) - M3*cos(theta3))/M4
b) x4 = -(525*12.02 + 425*16.97 - 325*13.4)/875
= -10.5 cm
c) simillarly,
y4 = -(M1*R1*sin(theta1) - M2*R2*sin(theta2) + M3*sin(theta3))/M4
d)
y4 = -(525*12.02 - 425*16.97 + 325*13.4)/875
= -3.94 cm
e) R4 = sqrt(x4^2 + y4^2)
= sqrt(10.5^2 + 3.94^2)
= 11.2 cm
f) theta = tan^-1(y4/x4)
= tan^-1(-3.94/10.5)
= 20.6 degrees below -x axis
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.