Question

A swing is made from a rope that will tolerate a maximum tension of 4.95 x 10² N without breaking. Initially, the swing hangs vertically. The swing is then pulled back at an angle of 35.6 ° with respect to the vertical and released from rest. What is the mass of the heaviest person who can ride the swing?

Answer 1

Since rope will have maximum tension ay bottom of swing.

So, by force balance on person at the bottom of swing,

T = m*g + m*v^2/R eq(1)

here, T = maximum tension in rope = 4.95*10^2 N = 495 N

m = mass of person = ??

g = surface gravity = 9.81 m/s^2

R = length of rope

v = speed of person at bottom = ??

using energy conservation,

KEi +PEi = KEf + PEf

here, KEi = 0 (As, person released from rest)

PEi = m*g*h

KEf = 0.5*m*v^2

PEf = 0

h = initial height of person = R - R*cos

= released angle from vertical = 35.6 deg

So, 0 + 0.5*m*v^2 = m*g*h + 0

v = sqrt(2*g*R*(1 - cos))

now, from equation 1,

T = m*g + m*(2*g*R*(1 - cos))/R

T = m*g + m*2*g*(1 - cos)

m = T/(g + 2*g*(1 - cos))

Using known values:

m = 495/(9.81 + 2*9.81*(1 - cos(35.6 deg))) = 36.72934

m = 36.7 kg

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