Part One. A conveyor belt system delivers each 10kg crate filled with plastic do
ID: 1338126 • Letter: P
Question
Part One. A conveyor belt system delivers each 10kg crate filled with plastic dolls to the ramp at point A, such that the crate's velocity is 2.0m/s, directed down the ramp. If the coefficient of Kinetic friction between the crate and the ramp is .3, determine the speed at which each crate slides off the ramp at B and into the warehouse cart. Distance between A ( beginning of conveyer belt) and B (warehouse cart) is 3m. The angle is 30degrees.
Part B. The manufactur of the plastic dolls has changed the production line that this conveyor belt system services. Using the results from market research, the manufacturer is changing from producing plastic dolls to glass dolls. Your team has been given the task of deternining what changes need to be made to the current conveyor belt system. Tests has shown that the crates must have a speed of no more than 3 m/s when the crates reach the warehouse cart to prevent breakage. The team decides the best cost effective method would be to change the length of the ramp taking the crates from the conveyer belt to the warehouse cart. The cart will be redesigned by another team after you find the new dimensions. Using Part One and the information given on the diagram, determine the new length of the ramp and the new angle it forms with the horizontal.
Explanation / Answer
Frictional force Ff = u mg cos theta ,
using energy conservation energy lost UA-B = Ff = u mg cos theta
The box begins at a position a/2 above point B.
You are conservation of energy is then 0.5 mVa^2 + 0 .5 mga = 0.5 mVB^2
As you can see, all of the masses cancel, so you find that the velocity at point B
Vb = sqrt (vA^2 + g a + 2a u g cos(30)
Vb = sqrt *( 4 + (9.8 * 3 * 0.3* 9.8 * 0.866))
VB = 8.87 m/s
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for new length S = (0.5 m Va^2-Vb^2)/ mg
S= 0.5* (8.87^2 - 3^2)/(9.8)
S =3.554 m