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kirza4 [7]
3 years ago
15

A purple beam is hinged to a wall to hold up a blue sign. The beam has a mass of mb = 6.7 kg and the sign has a mass of ms = 17.

2 kg. The length of the beam is L = 2.78 m. The sign is attached at the very end of the beam, but the horizontal wire holding up the beam is attached 2/3 of the way to the end of the beam. The angle the wire makes with the beam is θ = 33.9°.
What is the tension in the wire? What is the net force the hinge exerts on the beam? The maximum tension the wire can have without breaking is

T=936 N. What is the maximum mass sign that can be hung from the beam?

Physics
1 answer:
-BARSIC- [3]3 years ago
5 0

Answer:

wire tension =1003.8N

force in the hinge=894.15N

mass of sign=15.8Kg

Explanation:

Hello!

To solve this problem you must follow the following steps the complete procedure is in the attached images

1. draw the complete free body diagram of the situation

2. Raise Newton's equilibrium equations that establish that for a body to be in equilibrium, the sums of its vertical, horizontal forces and moments with respect to point A must be equal to zero.

3. Use the moment equilibrium equation with respect to point A (see attached image) to find the cable tension.

4. Use the equilibrium equations of vertical and horizontal forces to find the total reaction force on the hinge.

5. Use the equilibrium equation of moments with respect to point A (see attached image) to find the mass of the sign.

Remember to use algebra correctly to find the variables.

Greetings!

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Suppose we wrap a string around the surface of a uniform cylinder of radius 38.0 cm that is supported by an axle passing through
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Answer:

Angular acceleration = 23.68 rad / s²

Explanation:

Given that,

acceleration = 9m/s²

Therefore acceleration of string is 9m/s²

since string is constant in length

cylinder of radius 38.0 cm = 0.38m

Angular acceleration = a / r

Angular acceleration = 9 / 0.38

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4 0
3 years ago
A wave travels at 175 m/s along the x-axis.If the period of the periodic vibrations of the wave is 3.00 milliseconds,then what i
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Answer:  

<h2>E) 52.5 cm</h2>

Explanation:

Step one:

given data

period T= 3 milliseconds= 0.003

velocity v= 175m/s

wave lenght λ=?

Step two:

we know that f=1/T

the expression relating period and wave lenght is

v=λ/T

λ=v*T

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7 0
2 years ago
On a frictionless horizontal air table, puck A (with mass 0.254 kg ) is moving toward puck B (with mass 0.367 kg ), which is ini
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Answer:

v_a=0.8176 m/s

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Explanation:

According to the law of conservation of linear momentum, the total momentum of both pucks won't be changed regardless of their interaction if no external forces are acting on the system.

Being m_a and m_b the masses of pucks a and b respectively, the initial momentum of the system is

M_1=m_av_a+m_bv_b

Since b is initially at rest

M_1=m_av_a

After the collision and being v'_a and v'_b the respective velocities, the total momentum is

M_2=m_av'_a+m_bv'_b

Both momentums are equal, thus

m_av_a=m_av'_a+m_bv'_b

Solving for v_a

v_a=\frac{m_av'_a+m_bv'_b}{m_a}

v_a=\frac{0.254Kg\times (-0.123 m/s)+0.367Kg (0.651m/s)}{0.254Kg}

v_a=0.8176 m/s

The initial kinetic energy can be found as (provided puck b is at rest)

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The final kinetic energy is

K_2=\frac{1}{2}m_av_a'^2+\frac{1}{2}m_bv_b'^2

K_2=\frac{1}{2}0.254Kg (-0.123m/s)^2+\frac{1}{2}0.367Kg (0.651m/s)^2=0.07969 J

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\Delta K=0.07969 J - 0.0849 J = -0.00521 J

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Explanation:

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