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A person pushes on the handle of a lawnmower with a force of 280. n. if the handle makes an angle of 40.0 degrees with the ground, calculate the coefficient of friction if the lawnmower weighs 350. n and is moving at a constant velocity.

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Kalahira
Force F = 280 N Angle with the ground = 40 degrees Weight of the Lawnmower = 350 N Velocity is constant so Acceleration is 0 So Forward force Ff = F cos theta = 280 cos40 Frictional force with resists to back Fb = (u x Force from pressure) + vertical component of Force, where u is the coefficient of friction. Fb = (u x m x g) + (u x 280sin40) AS Ff = Fb => 280 cos40 = u x ((m x g) + 280sin40) u = 280 cos40 / ((350 x 9.81) + 280sin40) = 214.49 / () = 0.405 So the coefficient of friction u = 0.405
Khoso123

Answer:

µ= 0.405

Explanation:

Acceleration is 0 (constant V), so all the forces in the horizontal direction must add up to 0. Specifically those are these two:

Horizontal component of the force of (F1). It acts FORWARD.  

F1 = F*cosθ = 280cos40 N

Force of friction

F(fr) = µ*(total pressure force).

Two forces are creating the pressure force (Fp).

The other is the vertical component of F It's equal to

F*sinθ = 280sin40.

Then the friction force

F(fr) = µ*(mg + 280sin40)

Since

F1 = F(fr),

280cos40 = µ*(mg + 280sin40)

µ = 280cos40 / (mg + 280sin40)

µ= 0.405

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