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3 years ago

5

A block of mass 10 kg and measuring 250 mm on each edge is pulled up an inclined surface on which there is a film of SAE 10W-30

oil at 30F (the oil film is 0.025 mm thick). Find the steady speed of the block if it is released. If a force of 75 N is applied to pull the block up the incline, find the steady speed of the block. If the force is now applied to push the block down the incline, find the steady speed of the block. Assume the velocity distribution in the oil film is linear. The surface is inclined at an angle of 30 from the horizontal.

1 answer:

Ivan3 years ago

5 0

Answer:

a) 2.53 * 10^-2 m/s

b) -4.78 * 10^-2 m/s

c) 1.21 * 10^-1 m/s

Explanation:

Given data :

Mass of block = 10 kg

Measuring 250mm on each side

a) calculate the speed when a force of 75N is applied to pull block upwards

F = f + W sin∅ ( equation for applying the force of equilibrium condition in the x axis ) ----- ( 1 )

f ( friction force )= ( 16400v * 6.25 *10^-2) = 1025 v

F ( force applied ) = 75

W ( weight of block ) = 10 * 9.81 = 98.1 N

∅ = 30°

input values into equation 1

V = $\frac{75- (98.1*sin30^{0}) }{1025}$ = 2.53 * 10^-2 m/s

b) Speed when no force is applied on the block

F = f + W sin∅

F = 0

f = 1025 V

W = 98.1 N

∅ = 30°

hence V = $\frac{0 - (98.1*sin30^{0}) }{1025}$ = - 4.78 * 10^-2 m/s

c) when a force is applied to push block down the incline

F = f + W sin∅ ----- ( 3 )

F = 75 N

f = 1025 V

W = 98.1 N

∅ = 30°

input values into equation 3 considering the fact that the weight of the block is acting in the opposite direction

75 = 1025 V - 98.1 ( sin 30° )

V = $\frac{75+( 98.1*sin30^{0}) }{1025}$ = 1.21 * 10^-1 m/s

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