A Belt Drive Has Two Pulleys Of Radius R.1 And R 2.. Chegg.com E Chegg Study TEX

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A Belt Drive Has Two Pulleys Of Radius R.1 And R 2.. Chegg.com E Chegg Study TEXTBOOK SOLUTIONS EXPERT OEA udy / engineering / mechanical engineering /mechanical engineering questions and answers /a belt drive has two pulleys Question: A belt drive has two pulleys of radius r_1 and r_2 with a 4.3 A belt drive has two pulleys of radius r, and r, with a rubber belt passing around them (Figure P4.3). For tightening the belt, one pulley can be moved slightly, center-to-center using a slider mechanism with tightening screws. Suppose that 2r.-0.25 m and 2,-0.35 m. Initially, when the belt was not stretched (i.e., it was unstrained), the center-to-center distance between the pulleys was a,-1.0 m. Next, the belt was tightened by increasing this distance to a= 1.02 m. Estimate the average normal strain of the belt in this configuration. A rigid platform of length L( 10 m) is hinged at one end and held horizontally using a verti cal cable of length b5 m), which is attached to the platform at a distance a ( 6 m) from 44 Mg FIGURE P4.2 A horizontal cable carrying a load. Rubber belt Pulley with 14

Explanation / Answer

4.3. given 2r1 = 0.25 m

2r2 = 0.35 m

initial belt to belt distance ao = 1 m

new belt to belt distance a = 1.02 m

now, angle between the line joioning the centers of the pulleys ant the belt is theta

sin(theta) = r1/x = r2/(x + a)

r1*(x + a) = r2*x

x = r1*a/(r2 - r1)

sin(theta) = (r2 - r1)/a

let initial length of the belt between the pulleys be l1

then tan(theta1) = r1/y = r2/(y + l1) = sin(theta1)/sqroot(1 - sin^2(theta1))

r1(y + l1) = r2*y

y = r1*l1/(r2 - r1)

and tan(theta1) = (r2 - r1)/l1 = (r2 - r1)/sqroot(ao^2 - (r2 - r1)^2)

l1 = sqroot(ao^2 - (r2 - r1)^2)

l1 = sqroot(1^2 - (0.05)^2) = 0.998749

similiarly after strech, l2 = sqroot(a^2 - (r2 - r1)^2)

l2 = sqroot(1.02^2 - (0.05)^2) = 1.018773

total length of the belt originally = 2l1 + r1*(180 - 2*theta1)*pi/180 + r2(180 + 2*theta1)*pi/180

tan(theta1) = (0.175 - 0.125)/0.998794

theta1 = 2.86585

L = 2*0.99874 + 0.125(180 - 2*2.86585)*pi/180 + 0.175(180 + 2*2.86585)*pi/180 = 2.94495 m

and dL = 2*(dl2 - dl1) = 0.040048

average strain = dL/L = 0.01359887

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