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A ROBOT UNLOADS A FINISHED COMPONENT FROM A
MACHINE AND PLACES IT ON A GRAVITY CONVEYOR. THE CONVEYOR ACCELERATES FROM REST
UNTIL IT REACHES THE BOTTOM OF THE CONVEYOR 4.5 SECONDS LATER. THE COMPONENT IS
TRAVELLING AT A VELOCITY OF 1.35 M/S WHEN IT REACHES THE BOTTOM OF THE CONVEYOR
WHERE , IT IS THEN STOPPED BY HITTING A BUFFER. THE COMPONENT SITS STATIONARY
AWAITING PACKAGING.
A) DETERMINE THE LINEAR ACCELERATION
OF THE COMPONENT AS IT TRAVELS DOWN THE GRAVITY CONVEYOR (IGNORING FRICTIONAL
FORCES.
B) THE CONVEYOR SECTION LENGTH IS
LATER MODIFIED AND EXTENDED BY A FURTHER 1.5M. DETERMINE THE NEW VELOCITY OF
THE COMPONENTS WHEN THEY REACH THE BUFFER.
C) EXPLAIN THE CAUSE OF THE
COMPONENTS ACCELERATING FROM REST DOWN THE CONVEYOR.
D) DETERMINE THE FORCE EXERTED BY THE
COMPONENT, OF MASS 4.5KG, DOWNWARD ON THE CONVEYOR AS IT SITS AT THE
BUFFER.
E) EXPLAIN WHY THE COMPONENT DOES NOT
ACCELERATE DOWN THROUGH THE CONVEYOR WHILST RESTING AT THE BUFFER.
A)
FOR THE FIRST PART, USE EQUATION VF
= V0 + AT, WHERE V0 = 0, T = 4.5 S AND VF =
1.35 M/S
A = 1.35/4.5 = 0.3 M/S2B)
KNOWING ACCELERATION AND NEW DISTANCE, YOU
CAN DETERMINE THE NEW FINAL SPEED USING EQUATION:
(VF)2 = V02 + 2A*D
THE PROBLEM STATES THAT IT IS EXTENDED BY AN ADDITIONAL 1.5 M, SO YOU’LL HAVE
TO FIRST DETERMINE THE INITIAL LENGTH BEFORE THE EXTENSION:
D = VF2/2A = (1.35)^2/(2*.3) = 3.0375 M
NOW YOU ADD 1.5 TO THAT DISTANCE ABOVE AND USE THE FORMULA TO DETERMINE VF:
VF = ROOT(2*.3*4.5375) = 1.65M/S
C) GRAVITY
D)
WE KNOW THAT
VELOCITY (V) = ACCELERATION (A) X TIME (T)
1.35 = A X 4.5 ———à A=0.3 M/S2
FORCE (F) = MASS (M) *ACCELERATION (A)
=
4.5 KG *0.3 M/S2 =1.35
E)
THE FORCE EXERTED WILL BE EQUAL TO MASS
TIMES THE GRAVITY. THAT IS 4.5*9.81= 44.145 N.
THE COMPONENT DOES NOT ACCELERATE DOWN THE
CONVEYOR BELT BECAUSE WHEN ENTIRE ASSEMBLY IS CONSIDERED AS A SYSTEM, IT DOES
NOT AFFECT OUR SOLUTION AND AS IT DOES NOT AFFECT THE SOLUTION. THE
ACCELERATION OF ALL PARTS IS SAME AND HENCE, COMPONENT MAINTAINS ITS INERTIA
UNDER THE EFFECT OF GRAVITY

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