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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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ahrayia [7]

Answer:

distance from speaker is 17.87 m

Explanation:

given data

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what distances from the speaker

solution

we know sound velocity c = 331.5  + 0.6 × 20c = 343.5

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wavelength = 343.5 /  686 = 0.5 m

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d = λ/2 × (2n-1)

for n = 1, 2 3 4 ..

d = 0.5/2 × (2n-1)

d = 0.250 , 0.75 , 1.25 , 1.750............   for n = 1, 2 3 .............

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for d = 0.250

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\sqrt{3^{2}+(2..5+x)^{2} } - (2.5 + x1) = 0.250

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\sqrt{3^{2}+(2..5+x)^{2} } - (2.5 + x2) = 0.75

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side of triangle by hypotenuse of triangle are

\sqrt{3^{2}+(2..5+x)^{2} } - (2.5 + x3) = 1.250

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x4 will be negative so we stop here

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distance = 2.5 + x

here x = 0.475 , 3.125 and 15.375 so

distance 1 = 2.5 + 0.475  = 2.975 m

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8 0
4 years ago
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Andrews [41]

Answer:

Explanation:

Hi there,

To get started, recall the kinematic equations from either a textbook, equation sheet, etc. Kinematic equations are used when acceleration is <em>constant,</em> as stated in the prompt.

Best way to use kinematic equations is to see which variable you are looking for, then which variable is unknown to you and is not needed for that equation.

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We can still use the average velocity formula, as the other two equations that include final velocity have acceleration variable which is unknown as of now.

Solve for final velocity:

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a = \frac{v-v_0}{t_t_o_t_a_l} = \frac{(74-8)m/s}{40s} =1.65 m/s^{2}

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Study well and persevere. If you liked this solution, hit Thanks or give a rating!

thanks,

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4 years ago
The force needed to keep a car from skidding on a curve varies inversely as the radius of the curve and jointly as the weight of
motikmotik

Explanation:

It is given that, the force needed to keep a car from skidding on a curve varies inversely as the radius of the curve and jointly as the weight of the car and the square of the car's speed such that,

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F=\dfrac{kmgv^2}{r}

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v is the speed of the car

Case 1.

F = 640 pounds

Weight of the car, W = mg = 2600 pound

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640=\dfrac{k(2600)(40)^2}{650}

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Hence, this is the required solution.

6 0
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