Minimum Force (roller over step)

[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.

 

[haruspex]

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

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)?