(a) Explanation Given information: The conveyor belt starts from rest and travels a distance ( x max ) of 1.2 ft . The acceleration ( a max ) of the train is limited to 4 ft/s 2 . The rate of change of acceleration ( a max Δ t ) is limited to 4.8 ft/s 2 per second. Calculation: At time ( t ) zero second the position ( x ) and velocity ( v ) is zero. The velocity is zero at the maximum distance ( x max ) of 1.2 ft . The maximum velocity ( v max ) occurs at time ( t ) 1 2 t min . Show the a-t curve for a conveyor belt as in Figure (1). Calculate the velocity ( v 1 ) at time Δ t using the relation: v 1 = v 0 + 1 2 ( Δ t ) ( a max ) Here, v 0 is initial velocity. Substitute 0 for v 0 . v 1 = 0 + 1 2 ( Δ t ) ( a max ) = 1 2 ( Δ t ) ( a max ) Calculate the velocity ( v max ) at time 2 Δ t using the relation: v max = v 1 + 1 2 ( Δ t ) ( a max ) Substitute 1 2 ( Δ t ) ( a max ) for v 1 . v max = 1 2 ( Δ t ) ( a max ) + 1 2 ( Δ t ) ( a max ) = ( Δ t ) ( a max ) Show the v-t curve for a conveyor belt using symmetry as in Figure (2). Here the area ( A 1 ) is equal to area ( A 2 ) and area ( A 3 ) is equal to area ( A 4 ) . so consider the shaded area. Calculate the shaded area ( A shaded ) using the relation: A shaded = v max ( 3 Δ t − Δ t ) = v max 2 Δ t Equate the shaded area A shaded to ( x max ) (b) To determine The maximum velocity ( v max ) and average velocity ( v ave ) of the belt during the time ( t min ) .

12th Edition

ISBN: 9781260265453

Author: BEER

Publisher: MCG

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Chapter 11.3, Problem 11.80P

(a)

To determine

The shortest time (tmin) required for the belt to move 12 ft.

(b)

To determine

The maximum velocity (vmax) and average velocity (vave) of the belt during the time (tmin).

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Chapter 11.3, Problem 11.79P

Chapter 11.3, Problem 11.81P

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The shortest time ( t min ) required for the belt to move 12 ft .

Solution Summary: The author shows the a-t curve for the conveyor belt as in Figure (1).