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# Physics Moments question?

A wheel of radius 0.50 m rests on a level road at point C and makes contact with the edge E of a kerb of height 0.20 m.

A horizontal force of 240 N, applied through the axle of the wheel at X, is required just to move the wheel over the kerb.

By taking moments about an appropriate point, show that the weight of the wheel is 180 N

A diagram can be seen on page 31 question 13 c on this webpage

### 2 Answers

- Steve4PhysicsLv 78 years agoFavorite Answer
Hard to explain without a diagram, so follow these steps.

Draw the diagram and add the weight of the wheel, W, downwards from X.

Draw a vertical line upwards from E so it meets the horizontal force-line at point P. Mark the length of EP = 0.5-0.2 = 0.3m.

Join X to E so you have a right-angled triangle, XEP.

From Pythagoras, 0.5² = XP² + 0.3². This gives XP = 0.4m (XEP is a '3-4-5' triangle).

Tale moments about P. On the point of turning, the clockwise moment from the 240N force = anticlockwise moment from W:

240 x 0.3 = W x 0.4

W = 240 x 0.3 / 0.4

W = 180N

If you are not sure about moment calculations try the video-lesson in the link.

Source(s): http://www.youtube.com/watch?v=ikB1SVvZTvU - Anonymous5 years ago
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