A cube of mass 22.05 lb with sides measuring 11.81 in\' in length, slides on a f

Question

A cube of mass 22.05 lb with sides measuring 11.81 in' in length, slides on a film of SAE 20W oil at 77 degree F having a film thickness of 0.002 in' (neglect edge effects). Calculate the tension in the rope if the velocity of the cube is constant at 2.297 ft/s Calculate the tension in the rope if the surfaces were dry (no oil) and the coefficient of sliding friction was 0.5. Would the tension for part a) be the same if the cube was accelerating? List the information used in part a) that was not required for part b)?

Explanation / Answer

here for SAE20W , viscosity = 90 to 180 centistokes at 77F.
let us assume viscosity to be (90+180)/2 = 135 cSt at 77F
= 135 x 1.076x10^-5 ft2/s = 0.209 in2/s

also density of  SAE20W = 0.888 g/cc = 888kg/m^3 = 888 x  0.000036127 lb/in3 = 0.02808 lb/in3

of  SAE20W =  0.209 in2/s x  0.02808 lb/in3 = 0.0058697 lb/in. s

T =  .A. du/dy = 0.0058697 x 11.812 x 2.297*12 / 0.002

here du = 2.297*12 -0 in.s = 2.297*12 in/s

and dy = 0.002 in

T = 112.83 KN

b) for dry friction = 0.5

T =  .A. du/dy = 0.5 x 11.812 x 2.297*12 / 0.002 = 961.129 KN

c) No,had there been acceleration the equation would have been

T - Frictional force = M x acceleration

since in qustn (a) acceleration = 0 T =  Frictional force

and if the body accelerates, T = M x acceleration +  Frictional force

T varies in both the cases

d) IN (a) we had to use the value of dynamic viscosity , of SAE20W, which was unknown to us

this was not required in part (c) as dry friction was only present.

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