Why do I keep getting this message? everyone else who has this question put in t

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Why do I keep getting this message? everyone else who has this question put in the same answer. I attached a link with the same question in another forum.

http://www.chegg.com/homework-help/questions-and-answers/part-machine-part-loaded-supported-shown-pass-theoretical-cutting-plane-machine-part-horiz-q4243970

Why do I keep getting this message? everyone else who has this question put in the same answer. I attached a link with the same question in another forum. A machine part is loaded and supported as shown. Pass a theoretical cutting plane through the machine part, horizontally, at K. Draw the top part as a free-body diagram, and calculate and show the applied force's x and y components at C. Note that we have chosen the top part because it is easier to use to solve for the internal reactions. Use the vector drawing tool to draw the components of the applied force at C (begin vector drawing at C). Next, draw the internal force and moment reactions that must exist to maintain static equilibrium (i.e., in the direction to balance the applied loads at C) on the section at K. Start at the black dots marked C and K. To add a moment, click twice on the dot marked M.

Explanation / Answer

you are doing wrong my friend!

where you cut the section there you have to add the force and moments , the situation of the point C will be same after cutting too because you are not applied or removed any forces at the point C

break the forces at C as Fcos60 in +ve y-direction

and Fsin60 in -ve x-direction

so the section CK to be in the equuillibrium like it was earlier before cutting all the force and moment will be balance

so the forces on K are

Fcos60 in negative y-direction to balance the force in y - direction

Fsin60 in positive x-direction to balance the force in x- direction

and now if you visualise the four forces , you can feel that the section CK rotates so there is unbalanced moment

to balance it we have to apply the moment with same magnitude but in opposite direction

so moment magnitude = Fsin60*75*10-3 = 175.37

and its direction is opposite into the plane of section and it applied on position K

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