The youngling plant in positive, PA is re-designing its materials handling equip
ID: 2995414 • Letter: T
Question
The youngling plant in positive, PA is re-designing its materials handling equipment. Figure 1 shows one of the newly introduced conveyer belts to be used to transport bottled cases (average weight of 8 kg) down the ramp to a loading dock. The coefficient of kinetic friction between the case and the ramp is muk = 0.25. If the crate starts from rest at A and the hydraulic device B exerts a Stopping force on the cases given as F = 180(1 + 0.15t)N, where t is in seconds measured from the time of first contact with the device, use Force-Acceleration method to determine the time t required to bring the case to rest.Explanation / Answer
Let velocity of crate as it reaches B be V
Force balance equation
N = m*g*cos30 = 8*9.81*0.866 = 67.96 N
Along the incline,
m*g*sin30 - mu*m*g*cos30 = m*a
a = 4.905 - 2.123 = 2.78 m/s2
v2 - u2 = 2*a*s, u = 0
v = 3.33 m/s
F = m*a = m* dv/dt
Along with time varying force friction would also stop the block
F_friction = mu*m*g*cos30 = 17 N
180 ( 1 + 0.15t) + 17 = 8 * - dv/dt
Integrating the above equation
22.5 [ t + 0.15 * t2/2] + 2.125 t= - v
Using limits as t ( 0 ---> t) , v ( 3.33 ----> 0)
22.5 [t + 0.15 * t2/2] + 2.125t= 3.33
0.075t2 + 1.0944t - 0.148 = 0
t = 0.134 s
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