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In the first figure here, a sled is held on an inclined plane by a cord pulling directly up the plane. The sled is to be on the verge of moving up the plane. In the second figure, the magnitude F required of the cord's force on the sled is plotted versus the coefficient of static friction μs between sled and plane: F1 = 2.2 N, F2 = 4.5 N, and μ2 = 0.512. At what angle θ (in ˚) is the plane inclined?

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meerkat18
When it says something like 'on the verge of moving,' it means that the pulling force and static friction force and gravitational force all cancel out! Any more pulling force and it is ready to move! At some point, you want F as a function of μs, to determine the force needed depending on the coefficient of static friction. This function, F(μs), will rely on the angle θ as well, but we want to consider just one angle θ in every scenario. One value means it is constant. But if we know the F, and we know μs, we can find what the constant angle θ must be! If F is the pulling force, FS is the static friction force, and FG is gravitational force, Fnet=0=F+FS+FG=F+FNμs+mgsinθ=F+mgcosθμs+mgsinθ=0 Then you can find F(μs), but then there is the issue of solving for the θ to make it true.

The equations of the forces acting when there is no friction and in the

presence of friction, gives a simultaneous equation in θ.

Response:

  • The angle at which the plane is inclined is approximately 26.1°

What are the relationships between friction and applied force on an inclined plane, and the angle of inclination?

F₁ = 2.2 N

F₂ = 4.5 N

μ₂ = 0.512

Force acting on an inclined plane, relationship equations are;

F = [tex]F_f[/tex] + m·g·sin(θ) = μ·[tex]\mathbf{F_N}[/tex] + m·g·sin(θ)

At F₁, μ = 0, which gives;

2.2 = m·g·sin(θ)

At F₂, μ₂ = 0.512, which gives;

4.5 = 0.512·m·g·cos(θ) + m·g·sin(θ)

Therefore;

[tex]m = \mathbf{\dfrac{2.2 }{9.81 \cdot sin(\theta)}}[/tex]

[tex]4.5 = 0.512 \cdot \dfrac{2.2 \cdot cos(\theta) }{9.81 \cdot sin(\theta)} \times 9.81 + \dfrac{2.2 \cdot 9.81 \cdot sin(\theta) }{9.81 \cdot sin(\theta)} = \mathbf{1.1264 \cdot cot(\theta) }+ 2.2}[/tex]

Which gives;

[tex]tan(\theta) = \dfrac{1.1264}{4.5 - 2.2} = \mathbf{\dfrac{1.1264}{2.3}}[/tex]

[tex]\theta = arctan\left(\dfrac{1.1264}{2.3} \right) \approx \mathbf{26.1^{\circ}}[/tex]

  • The plane is inclined at an angle θ ≈ 26.1°

Learn more about friction on an inclined plane here:

https://brainly.com/question/22051392

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