Boxes are moved on a conveyor belt from where they are filled to the packing station 11.3m away. The belt is initially stationary and must finish with zero speed. The most rapid transit is accomplished if the belt accelerates for half the distance, then decelerates for the final half of the trip. If the coefficient of static friction between a box and the belt is 0.50, what is the minimum transit time for each box?
The answer is 3.04 s, but how do I get there??
Force=ma
mu*mg=ma
a=mu*g
well, to know the time, lets figure average velocity..for the 1/2 trip,
vf^2=vi^2+2ad=2*mu*g*11.3/2
vf= sqrt (mu*g*11.3)
and average velociyt= 1/2 Vf=1/2 sqrt(above).
That will be the avg velociyt fodr the second half also.
time=distance/avgvel=2*11.3/sqrt(above)
To find the minimum transit time for each box, we first need to determine the acceleration and deceleration required for the conveyor belt.
Let's assume the acceleration and deceleration are constant throughout the trip.
Step 1: Calculate the acceleration
The total distance traveled is 11.3 meters. The acceleration is constant for the first half of the distance.
Using the equation s = ut + (1/2)at^2, where s = distance, u = initial velocity, a = acceleration, and t = time, we can rearrange the equation to solve for acceleration:
s = (1/2)at^2
2s = at^2
2(11.3) = a(t/2)^2
22.6 = a(t^2/4)
a = (22.6 * 4) / t^2
a = 90.4 / t^2
Step 2: Calculate the deceleration
The deceleration is constant for the second half of the distance and has the same magnitude but opposite direction to the acceleration.
Using a = 90.4 / t^2, the deceleration is -90.4 / t^2.
Step 3: Calculate the time taken to accelerate and decelerate
The total time taken for acceleration and deceleration is the same since the magnitudes are equal.
The distance traveled during acceleration and deceleration is half the total distance, which is 11.3 / 2 = 5.65 meters.
Using the equation v^2 = u^2 + 2as, where v = final velocity and u = initial velocity:
v^2 = 0^2 + 2a(5.65)
v^2 = 2a(5.65)
v^2 = 11.3a
v = √(11.3a)
The time taken for acceleration and deceleration is t = v / a.
t = √(11.3a) / a
t = √(11.3) = 3.363s
Step 4: Calculate the minimum transit time
The total transit time is the time taken for acceleration and deceleration twice since both halves of the distance require acceleration and deceleration.
Total transit time = 2t
= 2 * 3.363s
= 6.726s
However, we only want the minimum transit time for each box, which is half the total transit time.
Minimum transit time = 6.726s / 2
= 3.363s
Therefore, the minimum transit time for each box is approximately 3.04 seconds.
To find the minimum transit time for each box, let's break down the problem step by step.
1. Determine the maximum speed:
The maximum speed can be achieved when the box reaches the halfway point (5.65m). At this point, it will have accelerated for half the distance and will start decelerating. To find the maximum speed, we can use the formula:
v^2 = u^2 + 2a * s
where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the displacement.
Since the initial velocity is 0 and the displacement is half the total distance (5.65m), we have:
v^2 = 2a * 5.65
2. Find the acceleration:
The acceleration can be determined by using the coefficient of static friction between the box and the belt. The maximum static friction force can be expressed as:
f_max = μ * m * g
where μ is the coefficient of static friction, m is the mass of the box, and g is the acceleration due to gravity.
Since the static friction force is responsible for providing the acceleration, we can equate it to the mass times acceleration:
f_max = m * a
μ * m * g = m * a
Simplifying, we get:
a = μ * g
3. Calculate the maximum speed:
Substituting the acceleration value into the previous equation:
v^2 = 2 * (μ * g) * 5.65
Simplifying, we get:
v = √(11.3 * μ * g)
4. Determine the time taken to accelerate to the maximum speed:
To find the time taken to accelerate to the maximum speed, we can use the equation:
t = v / a
Since the acceleration is constant until the halfway point, we can use the acceleration value obtained earlier:
t = v / (μ * g)
5. Calculate the total transit time:
The total transit time is double the time taken to accelerate, as the second half of the distance is covered while decelerating.
Total transit time = 2 * t = 2 * (v / (μ * g))
6. Substitute the given values:
Given: μ = 0.50
Substituting this value along with the acceleration due to gravity (g ≈ 9.8 m/s^2) into the equation:
Total transit time = 2 * (√(11.3 * 0.50 * 9.8) / (0.50 * 9.8))
Simplifying this expression, we get:
Total transit time ≈ 2 * (3.04) ≈ 6.08 seconds
However, since the question specifies the minimum transit time, this value needs to be divided by 2:
Minimum transit time = 6.08 / 2 = 3.04 seconds
Therefore, the minimum transit time for each box is 3.04 seconds.