A07. In a loading dock, grain falls from a vertical height of 5.00 m on to a con

Question

A07. In a loading dock, grain falls from a vertical height of 5.00 m on to a conveyor belt that moves horizontally to the
right at a constant speed of 3.50 m/s. 150 kg of grain land on the belt in each second.
(i) Calculate the velocity of the grain at the instant when it hits the belt.
(ii) Calculate the average change in the momentum of the grain in each second when it strikes the belt. Express your
answer as a vector in Cartesian component form.
(iii) Calculate the average force exerted by the belt on the grain. Be careful to specify its magnitude and direction

Explanation / Answer

( i )

when it hits the belt , its final velocity = ( 2 g h )

v = ( 2 * 9.8 * 5 ) = 9.8994 m /s ;

( i i )

p1 = 150 * 9.8994 ( -j ) ;

p2 = + 150 * 3.5   ( i );

change in momentum = p2 - p1 = 3.5 * 9.8994 ( i ) - 150 * 3.5 ( -j )

p= 34.64 i + 525 j ;

( iii )

the force on grain = p / 1s = 34.64 i + 525 j

force on belt = - [ 34.64 i + 525 j ] ;

magnitude = ( 34.64^2 + 525^2 ) = 526.142 N ;

direction = tan inv ( -525 / - 34.64 ) = 86.00

actual direction = 86.00 + 180 = 266.00 w r t right direction ( taken as posiitvie x axis ]

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