A possible design for a simple scale to weigh objects is shown. the length of th

Question

A possible design for a simple scale to weigh objects is shown. the length of the string AB is 0.5 m. When an object is placed in the pan, the spring stretches and the string AB rotates. the object's weight can be determined by observing the change in the angle alpha. (a) Assume that objects with masses in the range 0.2-2 kg are to be weighed. Choose the unstretched length and spring constant of the spring in order to obtain accurate readings for weights in the desired range. (Neglect the weights of the pun and spring. Notice that a significant change in the angle alpha is needed to determine the weight accurately.)

Explanation / Answer

Let the weight of the object be W

when there is no weith on the pan
length of spring = l
Length AB = 0.5 m
angle of AB with horizonrtal = alphao
angle pf spring weith horizontal = thetao
then l*cos(thetao) + 0.5*cos(alphao) = 1

now, after the pan is loaded with weight W
Tension in string = T
Tension in spring = T'
angles at this moment = theta, alpha

so, Tcos(alpha) = T'cos(theta)
Tsin(alpha) + T'sin(theta) = W

also, T' = kx
where x = 0.5*alpha*pi/180 [ approximately]

so, W = 0.5*k*alpha*pi*sin(theta)/180 + Tsin(alpha)
but T = T'cos(theta)/cos(alpha)
so W = k0.5*alpha*pi*sin(theta)/180 + k(0.5*alpha*pi)cos(theta)sin(alpha)/180*cos(alpha)
W = 0.0087266*k[alpha*sin(theta) + alpha*cos(theta)*sin(alpha)/cos(alpha)]
W = 0.0087266*k*alpha*[sin(theta) + cos(theta)*tan(alpha)]
for maximum Weight W = 2 kg = 2*9.8 = 19.62 N
2248.298 = k*alpha*[sin(theta) + cos(theta)*tan(alpha)]

if we assume trigonometric part to be 1 ( as the angles arent going to be too big or too small)
kmax = 2248.298

similiarly for W = 0.2*9.81 N
kmin = 224.8

hence k should rane between 200 - 2500 N/m

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