# Minimum Force (roller over step)

1. Nov 7, 2013

### BMcC

On an inclined plane (φ = 12°), calculate the minimum force P required to pull the roller over the step. The radius of the roller, r = 0.685m

My attempt

The sum of the Moment force about the system = 0

0 = mg(sin(φ)*(r-h) + mg(cos(φ)*√[r2-(r-h)2] - P(cos(θ)*r-h)

P(cos(θ)*r-h) = mg(sin(φ)*(r-h) + mg(cos(φ)*√[r2-(r-h)2]

P(cos(32.9)*0.575m) = (26kg)(9.8)(sin(12)*(0.575) + (26)(9.8)(cos(12)*√[0.6852 - (0.575)2]

0.483*P = 30.46 + 92.79

P = 255.17

This answer appears to be wrong. Any help on this? I can't seem to figure out where I went wrong.

2. Nov 7, 2013

### haruspex

I guess you meant $0 = mg\sin(φ)(r-h) + mg\cos(φ)\sqrt{r^2-(r-h)^2} - P\cos(θ)(r-h)$.
What about P sin(θ)?

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