7. A horizontal force of 810 N and a...

7. A horizontal force of 810 N and a counterclockwise couple-moment M are applied to the axle of pulley A, which is supported by a roller. The coefficient of static friction between the flat belt and each pulley is (mu = 0.4). The radius of each pulley is 40 mm. Determine the maximum value of M and the corresponding tensions in belt segments AB and AB'. Answers must be supported with relevant free-body diagrams and equilibrium equations.

Transcript

00:01 In this question we have given a horizontal that a horizontal force of 810 newton and a counterclockwise couple moment m are applied to an axel of pulley a and we have given the static friction and the radius of the pulley we have to determine the value maximum value of m and corresponding tension in bell segment ab and ab dash first of all i am drawing the diagram for this question so this is our given pulley excel of pulley and there is belt on this pulley these are the belt segment here and tension in lead tension in this segment ab is t a b and tension in this segment is t a b dash and it is rotating in counterclockwise direction and there is a force 810 newton is acting and there is a moment m first of all we have given data for this question that is force f is equal to to 810 newton and we have given the radius that is 40 m or we can say this is 0 .04 meter and new is 0 .4 now we know that for fully the ratio of tension in tight side upon slack side that is t a b upon t a b dash is equal to this is e to the power mu into angle alpha so so this is equal to, now alpha is equal to this is the 180 degree or we can say this is pi in radian.

02:02 This is the angle made by this belt over this pulley and this angle is 180 degree.

02:08 So finally we can see that the ratio of tension ab upon ab dash this is equal to e to the power this is 0 .4 into pi and this will come out to be from here tab is equal to this is 3 .511 t a b dash this is the equation first year now the second equation is that net horizontal force is 0 so we can say that the value of t a b plus t ab dash is equal to 810 this is our equation 2 so if we substitute the value of equation the value of t a b from equation one so we can say it's a substitute the value of t a b from equation one in equation two so we get this is 3 .511 t a b dash plus t a b dash is equal to 810 or we can say that the value of t a b dash is equal to 179 .56 newton and we get the value of tab from equation 1 this is equals to 3 .511 into 179 .569 this will come out to be 630 .46 newton so this is the tension ab and this is the tension in ab dash this is from equation 1 now we have to find the moment m so we can say that net moment for equilibrium net moment about a is equals to 0 so we can say that the moment and this is in counter clockwise direction and the moment due to the tension forces this is 630 .46 and this is in clockwise direction and the distance is equals to the radius that is 0 .04 and moment due to ab dash is in counterclockwise direction so this is plus 179 .569 that is tension into distance is 0 .04 and this is equals to 0...

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