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MissTica
3 years ago
6

A 150 kg uniform beam is attached to a vertical wall at one end and is supported by a vertical cable at the other end. Calculate

the magnitude of the tension in the wire if the angle between the beam and the horizontal is θ = 33°.
Physics
1 answer:
Serhud [2]3 years ago
8 0

Answer:

T = 1351 N

Explanation:

Weight = mg = 150 x 9.81 = 1471.5 N

Vertical component = T sin θ

Horizontal component = Tcos θ

Now, Let’s determine the clockwise and counter clockwise torques by letting the pivot point be at the end of the beam that is attached to the wall.

Thus, the weight of the beam will produce clockwise torque and the vertical component of the tension will produce counter clockwise torque.

At the weight of the beam is at its center;

Clockwise torque = WL/2

Counter clockwise torque = TL sinθ

Now, clockwise torque will be equal to anti clockwise torque and thus;

TL sinθ = WL/2

Thus;

T sin 33 = 1471.5/2

T sin 33 = 735.75

T = 735.75 ÷ sin 33 = 735.75/0.5446

T = 1350.99

This is approximately 1351 N.

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Answer:

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A ball is shot straight up into the air from the ground with initial velocity of 44ft/sec44ft/sec. assuming that the air resista
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Convert the given in SI units.

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The distance traveled and the initial velocity can be related through the equation,
  
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The value of d from the equation,

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If you haven't learned about velocity yet, just think about it as speed for now. The funny-looking triangle, ∆, is a symbol for "the change of." For example, if I started walking at 3 \frac{feet}{second} then sped up to 5 \frac{feet}{second}, then the change in my speed would be 2 \frac{feet}{second}, because I started walking 2 \frac{feet}{second} faster!

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Now, let's plug in our values! ∆v is the change in velocity, and to find that we simply have to subtract 16 \frac{meters}{second} by 4 \frac{meters}{second}. That makes sense, right? Back to the equation.

a = \frac{∆v}{t}
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We have our answer! The car's acceleration is 3 meters per second^{2}.

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