Hey, so I'm building a robot with a pivot flywheel. As of now the pivot is at 180 degrees but can be rotated to 90 degrees. I'm using air pumps to move it to the 90-degree angle, what gearing do you think would work best with it?
as per definitions ''In hinged beams, there are only two directions of motion being resisted - horizontal and vertical. There is, however, no resistance towards any rotation within the supports''. hence the moment should be zero at O. but we are getting non zero value. pls help me..........
In this example, there is 35º between the 'F' axle and the 'r' axle. Thus, in order to find the X component of the force 'F', there must be in horizontal. He just subtract the 20º angle from the 35º angle to have the horizontal. Then the result was a 15º angle in relation to the horizontal, or 0º degree angle.
Sir, since we have got the vector magnitude at the first step, can we just use the right thumb law or some others methods to find out the vector direction which is in the k direction, instead of doing the cross product?
The right thumb rule will give you the direction (in a graphical sense), but you'll need the cross product to represent it in an exact mathematical way.
It simply depends on what reference angle you are using. (Always use the definition: Cos(angle) = adjacent / hypothenuse and sin(angle) = opposite / hypothenuse )
Hey, so I'm building a robot with a pivot flywheel. As of now the pivot is at 180 degrees but can be rotated to 90 degrees. I'm using air pumps to move it to the 90-degree angle, what gearing do you think would work best with it?
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using the right hand rule, should the moment be going into the board?
Using the definition Torque = r x F the direction of the moment is positive (out of the board)
as per definitions
''In hinged beams, there are only two directions of motion being resisted - horizontal and vertical. There is, however, no resistance towards any rotation within the supports''.
hence the moment should be zero at O.
but we are getting non zero value.
pls help me..........
sir could you explain why we took 15 degrees for Fx and Fy at 6:10
In this example, there is 35º between the 'F' axle and the 'r' axle. Thus, in order to find the X component of the force 'F', there must be in horizontal. He just subtract the 20º angle from the 35º angle to have the horizontal. Then the result was a 15º angle in relation to the horizontal, or 0º degree angle.
What if there's a unit mm, should we convert it into m??
If you are not sure, then I recommend you convert to standard units.
Thank you
why the Fx and Fy component of the force cannot be in the positive direction?
The force acts towards the bottom left corner, thus the components are negative.
Sir, since we have got the vector magnitude at the first step, can we just use the right thumb law or some others methods to find out the vector direction which is in the k direction, instead of doing the cross product?
The right thumb rule will give you the direction (in a graphical sense), but you'll need the cross product to represent it in an exact mathematical way.
Thank you so much sir !
Thanks a lot!
You're welcome!
Why did you use sin when finding the magnitude?
It depends on which angle you use.
sir when should we use cos ?
It simply depends on what reference angle you are using. (Always use the definition: Cos(angle) = adjacent / hypothenuse and sin(angle) = opposite / hypothenuse )
Nice
Coordinate system was never defined in the diagram..
thank you sir