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AURORKA [14]
2 years ago
13

Beth, a construction worker, attempts to pull a stake out of the ground by pulling on a rope that is attached to the stake. The

rope makes an angle of 44.2°with the horizontal. If Beth exerts a force of 153.8 N on the rope, what is the magnitude of the upward component of the force acting on the stake? Answer in units of N.
Physics
1 answer:
AnnyKZ [126]2 years ago
8 0

Answer:

Fy=107.2 N

Explanation:

Conceptual analysis

For a right triangle :

sinβ = y/h formula (1)

cosβ = x/h formula (2)

x: side adjacent to the β angle

y:  opposite side of the β angle

h: hypotenuse

Known data

h = T = 153.8 N : rope tension

β= 44.2°with the horizontal (x)

Problem development

We apply the formula (1) to calculate Ty : vertical component of the rope force.

sin44.2° = Ty/153.8 N

Ty = (153.8 N ) *(sen44.2°)= 107.2 N  directed down

for equilibrium system

Fy= Ty=107.2 N

Fy=107.2 N upward component of the force acting on the stake

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At a depth of 1030 m in Lake Baikal (a fresh water lake in Siberia), the pressure has increased by 100 atmospheres (to about 107
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A volume of a cubic meter of water from the surface of the lake has been compressed in 0.004 cubic meters.

Explanation:

The bulk modulus is represented by the following differential equation:

K = - V\cdot \frac{dP}{dV}

Where:

K - Bulk module, measured in pascals.

V - Sample volume, measured in cubic meters.

P - Local pressure, measured in pascals.

Now, let suppose that bulk remains constant, so that differential equation can be reduced into a first-order linear non-homogeneous differential equation with separable variables:

-\frac{K \,dV}{V} = dP

This resultant expression is solved by definite integration and algebraic handling:

-K\int\limits^{V_{f}}_{V_{o}} {\frac{dV}{V} } = \int\limits^{P_{f}}_{P_{o}}\, dP

-K\cdot \ln \left |\frac{V_{f}}{V_{o}} \right| = P_{f} - P_{o}

\ln \left| \frac{V_{f}}{V_{o}} \right| = \frac{P_{o}-P_{f}}{K}

\frac{V_{f}}{V_{o}} = e^{\frac{P_{o}-P_{f}}{K} }

The final volume is predicted by:

V_{f} = V_{o}\cdot e^{\frac{P_{o}-P_{f}}{K} }

If V_{o} = 1\,m^{3}, P_{o} - P_{f} = -10132500\,Pa and K = 2.3\times 10^{9}\,Pa, then:

V_{f} = (1\,m^{3}) \cdot e^{\frac{-10.1325\times 10^{6}\,Pa}{2.3 \times 10^{9}\,Pa} }

V_{f} \approx 0.996\,m^{3}

Change in volume due to increasure on pressure is:

\Delta V = V_{o} - V_{f}

\Delta V = 1\,m^{3} - 0.996\,m^{3}

\Delta V = 0.004\,m^{3}

A volume of a cubic meter of water from the surface of the lake has been compressed in 0.004 cubic meters.

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3 years ago
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