Moment Direction: 560sin30 (0.25) Explained

[werson tan]

Homework Statement

why the moment 560sin30 (0.25) isn't in clockwise direction ? the horizontal component of the 560N force is directed to the left , causing the moment in clockwise direction, am i right ?
I have made my sketch beside in the diagram ...

Homework Equations

The Attempt at a Solution

 

[Herman Trivilino]

The point of application of the 560 lb force is 0.25 ft above the hinge pin, not below as you have drawn it.

 

[werson tan]

The point of application of the 560 lb force is 0.25 ft above the hinge pin, not below as you have drawn it.

Why it can't be below the 0.25ft??

 

[Herman Trivilino]

I don't understand your question. What I'm saying is, when I look at the figure you posted, I see that the point of application of the force is above the hinge pin.

Look at the figure. Draw a line parallel to the concrete shoot that passes through the hinge pin and call it Line P. Draw another line parallel to the first line, but passing through the point of application of the 560 lb force. Call it Line A. Those two lines are separated by a distance of 0.25 ft. Line P is above Line A. The sketch you made in the margin is consistent with Line A being above Line P. Your way means the moment of the 560 lb force is negative (clockwise). But in reality the force creates a positive moment. It will tend to make the chute rotate counter-clockwise.

 

[haruspex]

Why it can't be below the 0.25ft??

Take a line through G parallel to the chute. Follow it up to where it is near A. It passes above A, at a distance of 0.25 ft from A.
The component of the 560lb parallel to the chute acts through G, down the slope. The moment of it about A is therefore anticlockwise.

 

[werson tan]

I don't understand your question. What I'm saying is, when I look at the figure you posted, I see that the point of application of the force is above the hinge pin.

Look at the figure. Draw a line parallel to the concrete shoot that passes through the hinge pin and call it Line P. Draw another line parallel to the first line, but passing through the point of application of the 560 lb force. Call it Line A. Those two lines are separated by a distance of 0.25 ft. Line P is above Line A. The sketch you made in the margin is consistent with Line A being above Line P. Your way means the moment of the 560 lb force is negative (clockwise). But in reality the force creates a positive moment. It will tend to make the chute rotate counter-clockwise.

I said that is clockwise becoz I thought that the force is located below the 0.25m ...ok, I know my mistakes already