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alex41 [277]
1 year ago
5

A plank of length L=2.200 m and mass M=4.00 kg is suspended horizontally by a thin cable at one end and to a pivot on a wall at

the other end as shown. The cable is attached at a height H=1.70 m above the pivot and the plank's CM is located a distance d=0.700 m from the pivot.
Calculate the tension in the cable.

Physics
1 answer:
Elenna [48]1 year ago
3 0

The tension in the cable is 23.2 N

<h3>What is the tension in the string?</h3>

The tension in the cable can be resolved into horizontal and vertical forces Tcosθ and Tsinθ respectively.

Tcosθ, is acting perpendicularly, Tcosθ = 0

Taking moments about the pivot:

Tsinθ * 2.2 = 4 * 9.8 * 0.7

Solving for θ;

θ = tan⁻¹(1.4/2.2) = 32.5°

T = 27.44/(sin 32.5 * 2.2)

T = 23.2 N

In conclusion, the tension in the cable is determined by taking moments about the pivot.

Learn more about moments of forces at: brainly.com/question/23826701

#SPJ1

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A wheel has a radius of r = 2.0 m and it rolls down a smooth incline. The height of the incline is h = 8.0 m . What is the angul
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The angular velocity of the wheel at the bottom of the incline is 4.429 rad/sec

The angular velocity (ω) of an object is the rate at which the object's angle position is changing in relation to time.

For a wheel attached to an incline angle, the angular velocity can be computed by considering the conservation of energy theorem.

As such the total kinetic energy (K.E) and rotational kinetic energy (R.K.E) at a point is equal to the total potential energy (P.E) at the other point.

i.e.

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\mathbf{mgh = \dfrac{1}{2}m(r \times \omega)^2 + \dfrac{1}{2}\times I \times \omega^2}

\mathbf{gh = \dfrac{1}{2}(r \times \omega)^2 + \dfrac{1}{2}\times r^2 \times \omega^2}

\mathbf{2 \times \dfrac{gh}{r^2} =\omega^2 +  \omega^2}

\mathbf{2 \omega^2=2 \times \dfrac{9.81 \times 8 m }{2.0 ^2}  }

\mathbf{\omega^2=\dfrac{39.24 }{2}}

\mathbf{\omega=\sqrt{19.62 } \ rad/sec}

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