To determine the force exerted by a fluid Part A At a given point, a fluid at re

Question

To determine the force exerted by a fluid Part A At a given point, a fluid at rest exerts a pressure, p that is the same in all directions. The magnitude of p can be determined from the following relationship A gate rotates at point A and releases water when the water's depth, d, exceeds a certain value. (Figure 1) The gate's width (perpendicular to the image) is b 3.30 ft, the water's depth is d= 3.54 ft, and point A is located at h 1.77 ft what is F, the magnitude of the horizontal force exerted on the gate by the water? The water's specific weight is Y 62.4 lb/ft3 where is the fluid's mass density, is the fluid's specific weight, z is the point's depth (measured from the fluid's surface), and g is the acceleration due to gravity. This pressure creates a linear distributed load on a flat vertical or inclined surface and a uniform distributed load on a horizontal surface. The resultants of these loadings can be determined by finding the volume under the loading curves. Express your answer numerically with the appropriate units to three significant figures Hints F1290 lb Submit Give Up Correct Figure 1 Part B What are A and B, the horizontal reaction forces at points A and B that act on the gate from Part A? (Figure 1) Express your answers numerically in pounds to three significant figures separated by a comma Hints vec A, B,- lb Submit My Answers Give Up

Explanation / Answer

We know that effective application of pressure force of water is at d/3 from bottom.

Balancing torque about A,

F*(h-d/3) = Bx *h

Bx = F*(h-d/3)/h = 1290*(1.77-3.54/3)/1.77 = 430 lb,

Ax = F-Bx = 1290 -430 = 860 lb.

So Ax, Bx = 860 , 430 lb answer

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