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PilotLPTM [1.2K]
3 years ago
9

What is the torque τa about axis a due to the force f⃗ ? express the torque about axis a at cartesian coordinates (0,0)?

Physics
1 answer:
pentagon [3]3 years ago
5 0

Answer:

\tau_a = F a sin \theta

Explanation:

The torque of a force is given by:

\tau = F d sin \theta

where

F is the magnitude of the force

d is the distance between the point of application of the force and the centre of rotation of the system

\theta is the angle between the direction of the force and d

In this problem, we have:

F, the force

a, the distance of application of the force from the centre (0,0)

\theta, the angle between the direction of the force and a

Therefore, the torque is

\tau_a = F a sin \theta

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Three people pull simultaneously on a stubborn donkey. Jack pulls eastward with a force of 80.5 N, Jill pulls with 81.7 N in the
Gnesinka [82]

Answer:

F = 233.52 N,  θ' = 351.41º

Explanation:

In this exercise we must find the net force applied on the donkey.

For this we use Newton's second law, where we create a reference frame with the horizontal x axis

let's decompose the forces

Jack

        = 80.5 N

Jill

       cos 45 = F_{2x} / F₂2

       sin 45 = F_{2y} / F₂2

       F_{2x} = F₂ cos 45

       F_{2y} = F₂ sin 45

       F_{2x} = 81.7 cos 45 = 57.77 N

       F_{2y} = 81.7 sin 45 = 57.77 N

Jane

      cos (270 + 45) = F_{3x} / F₃3

      sin 315 = F_{3y} / F₃

      F_{3x} = 131 cos 315 = 92.63 N

      F_{3y} = 131 sin 315 = -92.63 N

the force can be found in each axis

X axis

         F_{x} = F_{1x} + F_{2x} + F_{3x}

         F_{x} = 80.5 +57.77 + 92.63

         F_{x} = 230.9 N

Axis y

         F_{y} = F_{1y} + F_{2y} + F_{3y}

         F_{y} = 0 + 57.77 -92.63

         F_{y} = -34.86 N

we can give the result in two ways

a) F = (230.9 i ^ - 34.86 j ^) N

b) in the form of module and angle

we use the Pythagorean theorem

         F = √(Fₓ² + F_{y}²

        F = √(230.9² + 34.86²)

        F = 233.52 N

let's use trigonometry for the angle

        tan θ = \frac{F_y}{F_x} }

        θ = tan⁻¹ (\frac{F_y}{F_x} })

        θ = tan⁻¹ (-34.86 / 230.9)

        θ = -8.59º

if we measure this angle from the positive side of the x-axis counterclockwise

          θ' = 360 -θ

          θ‘= 360- 8.59

          θ' = 351.41º

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3 years ago
Make the curvature radius 0.6 m, the refractive index 1.5, and the diameter 0.6 m. place the lamp so that the source of light is
vodka [1.7K]
We are given
r = 0.6 m
n = 1.5
D = 0.6 m, R1 = 30 cm
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We are asked to get the focal length and the distance of the focal plane from the lens

We use the formula
1 / f = ( n - 1) (1/R1 - 1/R2)
Substituting and solving for f
1/ f = (1.5 - 1) (1/30 - 1/120)
f = 80 cm

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