NCEES 2001 #113

[navyasw02]

I just worked #113 in the morning section of the 2001 NCEES sample test. It's got F1 = 700 and F2 = 100 pulling down on a belt driven pulley. Why is the torque (700-100)*radius instead of (700+100)*radius? After I got done scratching my head, I went back to Shigley's Machine Design and Example 3-9 on page 100 of the 8th edition shows basically the same setup and adds the two forces instead of subtracting.

 

[navyasw02]

I just worked #113 in the morning section of the 2001 NCEES sample test. It's got F1 = 700 and F2 = 100 pulling down on a belt driven pulley. Why is the torque (700-100)*radius instead of (700+100)*radius? After I got done scratching my head, I went back to Shigley's Machine Design and Example 3-9 on page 100 of the 8th edition shows basically the same setup and adds the two forces instead of subtracting.

Bueller?

 

[Bluedog]

I just worked #113 in the morning section of the 2001 NCEES sample test. It's got F1 = 700 and F2 = 100 pulling down on a belt driven pulley. Why is the torque (700-100)*radius instead of (700+100)*radius? After I got done scratching my head, I went back to Shigley's Machine Design and Example 3-9 on page 100 of the 8th edition shows basically the same setup and adds the two forces instead of subtracting.

The FBD of a pulley has tensions, one on the in the tight side and the other on the slack side acting in opposite directions to each other. The pulley tends to rotate in the direction of the tight side and the magnitude of total force causing it to turn is the difference of the tension in the tight side and the slack side.

If the pulley is attached to a suspended load, both tensions act in the direction of the load and hence the resultant force is the sum of two tensions.

 

[navyasw02]

I just worked #113 in the morning section of the 2001 NCEES sample test. It's got F1 = 700 and F2 = 100 pulling down on a belt driven pulley. Why is the torque (700-100)*radius instead of (700+100)*radius? After I got done scratching my head, I went back to Shigley's Machine Design and Example 3-9 on page 100 of the 8th edition shows basically the same setup and adds the two forces instead of subtracting.

The FBD of a pulley has tensions, one on the in the tight side and the other on the slack side acting in opposite directions to each other. The pulley tends to rotate in the direction of the tight side and the magnitude of total force causing it to turn is the difference of the tension in the tight side and the slack side.

If the pulley is attached to a suspended load, both tensions act in the direction of the load and hence the resultant force is the sum of two tensions.

That makes sense, thanks for the explanation.

 

[Lily]

Torque provoques rotation, so imagine these forces are equal, there will be no rotation then. However if you are calculating the bending moment, which is the ability of the shaft to twist around its main axis, then you will have to add the forces, because both forces contribute to the bending. Hope it's clear...