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tatiyna
3 years ago
10

A uniform steel girder of weight 22KN and of length 12m is lifted off the ground at one end by means of a crane. When the raised

end is 2.0m above the ground, the cable is vertical.
A. Sketch a free-body diagram of the girder in this position
b. Calculate the tension in the cable at this position and the force of the girder on the ground
Physics
1 answer:
Keith_Richards [23]3 years ago
8 0

Answer:

Explanation:

For free body diagram see attached sheet .

W is weight of steel girder acting at the middle point of its length . T is tension in the cable .

OB = √ ( 12² - 2² )

= 11.83 m .

OC = 11.83 / 2 = 5.915 m

Taking moment of tension T and weight W about point O

W x OC = T x OB

22 x 5.915 = T x 11.83

T = 22 x 5.915 / 11.83

= 11 kN

Considering forces acting in vertical direction and equating forces in opposite direction

T + R = W

R = W - T

= 22 - 11 = 11 KN

So force of grinder on the ground = R

= 11 KN.

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Answer:

The angle of elevation of the rocket is increasing at a rate of 48.780º per second.

Explanation:

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Where:

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y - Vertical distance between the rocket and the observer, measured in kilometers.

The angle of elevation of the rocket (\theta), measured in sexagesimal degrees, is defined by the following trigonometric relation:

\tan \theta = \frac{y}{x} (2)

If we know that x = 5\,km, then the expression is:

\tan \theta = \frac{y}{5}

And the rate of change of this angle is determined by derivatives:

\sec^{2}\theta \cdot \dot \theta = \frac{1}{5}\cdot \dot y

\frac{\dot \theta}{\cos^{2}\theta} = \frac{\dot y}{5}

\frac{\dot \theta\cdot (25+y^{2})}{25} = \frac{\dot y}{5}

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\dot \theta - Rate of change of the angle of elevation, measured in sexagesimal degrees.

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3 years ago
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Hi!
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