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murzikaleks [220]
3 years ago
8

A 50 kg crate is pulled across the ice by a rope that is angled at 30 degrees above the horizontal. If the tension in the rope i

s 100 newtons, what is the magnitude of the "lifting" and "pulling" components of the force in the rope?
Physics
1 answer:
denpristay [2]3 years ago
4 0
Let:
Vx = the pulling component of force
Vy = the lifting component of force

Vy:
Sin(n°) = Vy/hypotenuse
hypotenuse * Sin(n°) = Vy
100N*sin(30°) = Vy
50N = Vy

Vx:
Cos(n°) = Vx/hypotenuse
Hypotenuse * cos(n°) = Vx
100N*cos(30°) =Vx
about 86.6N = Vx
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A third point charge q3 is now positioned halfway between q1 and q2. The net force on q2 now has a magnitude of F2,net = 14.413
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Answer:

The value of  charge q₃ is 40.46 μC.

Explanation:

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Magnitude of net force F=14.413\ N

Suppose a point charge q₁ = -3 μC is located at the origin of a co-ordinate system. Another point charge q₂ = 7.7 μC is located along the x-axis at a distance x₂ = 8.2 cm from q₁. Charge q₂ is displaced a distance y₂ = 3.1 cm in the positive y-direction.

We need to calculate the distance

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r=\sqrt{x_{2}^2+y_{2}^2}

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r=\sqrt{(8.2\times10^{-2})^2+(3.1\times10^{-2})^2}

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Using formula of net force

F_{12}=kq_{2}(\dfrac{q_{3}}{r_{3}^2}+\dfrac{q_{1}}{r_{1}^2})

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14.413=9\times10^{9}\times7.7\times10^{-6}(\dfrac{q_{3}}{(0.0438)^2}+\dfrac{-3\times10^{-6}}{(0.0876)^2})

(\dfrac{q_{3}}{(4.38\times10^{-2})^2}+\dfrac{-3\times10^{-6}}{(0.0876)^2})=\dfrac{14.413}{9\times10^{9}\times7.7\times10^{-6}}

\dfrac{q_{3}}{(0.0438)^2}=207\times10^{-4}+3.909\times10^{-4}

q_{3}=0.0210909\times(0.0438)^2

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q_{3}=40.46\ \mu C

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