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help !! (1 Viewer)
- Thread starter Napoleon
- Start date
When 0 < θ < 90° then sinθ < tanθ
Hence √(rgsinθ ) <√(rgtanθ ), that is, it doesn't have enough speed so it will start to slip down to slope, hence friction will be up the slope.
Hence √(rgsinθ ) <√(rgtanθ ), that is, it doesn't have enough speed so it will start to slip down to slope, hence friction will be up the slope.
Dumsum
has a large Member;
nevermind.
For part ii)
∑ vertical forces = 0 hence -------> Ncosθ + Fsinθ = mg (1)
∑ horizontal forces = mv<sup>2</sup>/r -----> Nsinθ - Fcosθ = mv<sup>2</sup>/r (2)
sinθ x (1) - cosθ x (2) yeilds:
F(sin<sup>2</sup>θ + cos<sup>2</sup>θ ) = mgsinθ - mv<sup>2</sup>cosθ/r
subbing in the value of v
F = mgsinθ - mgsinθcosθ = mgsinθ(1 - cosθ ) up the hill
∑ vertical forces = 0 hence -------> Ncosθ + Fsinθ = mg (1)
∑ horizontal forces = mv<sup>2</sup>/r -----> Nsinθ - Fcosθ = mv<sup>2</sup>/r (2)
sinθ x (1) - cosθ x (2) yeilds:
F(sin<sup>2</sup>θ + cos<sup>2</sup>θ ) = mgsinθ - mv<sup>2</sup>cosθ/r
subbing in the value of v
F = mgsinθ - mgsinθcosθ = mgsinθ(1 - cosθ ) up the hill
Dumsum
has a large Member;
I pretty much had it until I read KFunk's post which proved me wrong (had wrong friction direction)
The trick though is in the line (1) x sin@ - (2) x cos@
The trick though is in the line (1) x sin@ - (2) x cos@