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GenaCL600 [577]
3 years ago
13

Use tug of war to explain balanced and unbalanced forces

Physics
1 answer:
Dimas [21]3 years ago
4 0
Forces equal the flag doesn't move either direction

Unequal forces the flag moves in the direction of the greater magnitude force
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If the moment acting on the cross section is M=630N⋅m, determine the maximum bending stress in the beam. Express your answer to
-BARSIC- [3]

Answer:

2.17 Mpa

Explanation:

The location of neutral axis from the top will be

\bar y=\frac {(240\times 25)\times \frac {25}{2}+2\times (20\times 150)\times (25+(\frac {150}{2}))}{(240\times 25)+2\times (20\times 150)}=56.25 mm

Moment of inertia from neutral axis will be given by \frac {bd^{3}}{12}+ ay^{2}

Therefore, moment of inertia will be

\frac {240\times 25^{3}}{12}+(240\times 25)\times (56.25-25/2)^{2}+2\times [\frac {20\times 150^{3}}{12}+(20\times 150)\times ((25+150/2)-56.25)^{2}]=34.5313\times 10^{6} mm^{4}}

Bending stress at top= \frac {630\times 10^{3}\times (175-56.25)}{34.5313\times 10^{6}}=2.1665127\approx 2.17 Mpa

Bending stress at bottom=\frac {630\times 10^{3}\times 56.25}{34.5313\times 10^{6}}=1.026242858\approx 1.03 Mpa

Comparing the two stresses, the maximum stress occurs at the bottom and is 2.17 Mpa

8 0
3 years ago
The velocity of sound apparatus is used in an investigation to determine the frequency of an unknown tuning fork. The temperatur
Hitman42 [59]

Answer:

Explanation:

The relationship of the speed of sound, its frequency, and wavelength is the same as for all waves: vw = fλ, where vw is the speed of sound, f is its frequency, and λ is its wavelength.

6 0
3 years ago
A tugboat tows a ship at a constant velocity. The tow harness consists of a single tow cable attached to the tugboat at point A
Y_Kistochka [10]

Answer:

The tensions in T_{BC} is approximately 4,934.2 lb and the tension in T_{BD} is approximately  6,035.7 lb

Explanation:

The given information are;

The angle formed by the two rope segments are;

The angle, Φ, formed by rope segment BC with the line AB extended to the center (midpoint) of the ship = 26.0°

The angle, θ, formed by rope segment BD with the line AB extended to the center (midpoint) of the ship = 21.0°

Therefore, we have;

The tension in rope segment BC = T_{BC}

The tension in rope segment BD = T_{BD}

The tension in rope segment AB = T_{AB} = Pulling force of tugboat = 1200 lb

By resolution of forces acting along the line A_F gives;

T_{BC} × cos(26.0°) + T_{BD} × cos(21.0°) = T_{AB} = 1200 lb

T_{BC} × cos(26.0°) + T_{BD} × cos(21.0°) = 1200 lb............(1)

Similarly, we have for equilibrium, the sum of the forces acting perpendicular to tow cable = 0, therefore, we have;

T_{BC} × sin(26.0°) + T_{BD} × sin(21.0°) = 0...........................(2)

Which gives;

T_{BC} × sin(26.0°) = - T_{BD} × sin(21.0°)

T_{BC} = - T_{BD} × sin(21.0°)/(sin(26.0°))  ≈ - T_{BD} × 0.8175

Substituting the value of, T_{BC}, in equation (1), gives;

- T_{BD} × 0.8175 × cos(26.0°) + T_{BD} × cos(21.0°) = 1200 lb

- T_{BD} × 0.7348  + T_{BD} ×0.9336 = 1200 lb

T_{BD} ×0.1988 = 1200 lb

T_{BD} ≈ 1200 lb/0.1988 = 6,035.6938 lb

T_{BD} ≈ 6,035.6938 lb

T_{BC} ≈ - T_{BD} × 0.8175 = 6,035.6938 × 0.8175 = -4934.1733 lb

T_{BC} ≈ -4934.1733 lb

From which we have;

The tensions in T_{BC} ≈ -4934.2 lb and  T_{BD} ≈ 6,035.7 lb.

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What is the conservation of energy?
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Energy cannot be created nor be destroyed
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