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monitta
3 years ago
9

4.116 The lid of a roof scuttle weighs 75 lb. It is hinged at corners A and B and maintained in the desired position by a rod CD

pivoted at C. A pin at end D of the rod fits into one of several holes drilled in the edge of the lid. For α 5 50°, determine (a) the magnitude of the force exerted by rod CD, (b) the reactions at the hinges. Assume that the hinge at B does not exert any axial thrust. Beer, Ferdinand. Vector Mechanics for Engineers: Statics (p. 219). McGraw-Hill Higher Education. Kindle Edition.

Engineering
1 answer:
babunello [35]3 years ago
5 0

Answer:

(a) The magnitude of force is 116.6 lb, as exerted by the rod CD

(b) The reaction at A is (-72.7j-38.1k) lb and at B it is (37.5j) lb.

Explanation:

Step by step working is shown in the images attached herewith.

For this given system, the coordinates are the following:

A(0, 0, 0)

B(26, 0, 0)

And the value of angle alpha is 20.95°

Hope that answers the question, have a great day!

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While discussing what affects the amount of pressure exerted by the brakes: Technician A says that the shorter the line, the mor
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Answer:

Only Technician B is right.

Explanation:

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Pressure applied on the pedal, P(pedal) = P(pad)

And the Pressure is the applied force/area for either pad or pedal. That is, P(pad) = Force(pad)/A(pad) & P(pedal) = F(pedal)/A(pedal)

If the area of piston increases, A(pad) increases and the P(pad) drops, Meaning, the pressure transmitted to the pad reduces. And for most cars, there's a pressure limit for the braking system to work.

If the A(pad) increases, P(pad) decreases and the braking force applied has to increase, to counter balance the dropping pressure and raise it.

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3 years ago
I need help on the Coderz Challenge missions 3 part 3. PLEASE HELP!
allsm [11]

Answer:

the answer how you analyzs the problwm

6 0
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The dam cross section is an equilateral triangle, with a side length, L, of 50 m. Its width into the paper, b, is 100 m. The dam
lisabon 2012 [21]

Answer:

Explanation:

In an equilateral trinagle the center of mass is at 1/3 of the height and horizontally centered.

We can consider that the weigth applies a torque of T = W*b/2 on the right corner, being W the weight and b the base of the triangle.

The weigth depends on the size and specific gravity.

W = 1/2 * b * h * L * SG

Then

Teq = 1/2 * b * h * L * SG * b / 2

Teq = 1/4 * b^2 * h * L * SG

The water would apply a torque of elements of pressure integrated over the area and multiplied by the height at which they are apllied:

T1 = \int\limits^h_0 {p(y) * sin(30) * L * (h-y)} \, dy

The term sin(30) is because of the slope of the wall

The pressure of water is:

p(y) = SGw * (h - y)

Then:

T1 = \int\limits^h_0 {SGw * (h-y) * sin(30) * L * (h-y)} \, dy

T1 = \int\limits^h_0 {SGw * sin(30) * L * (h-y)^2} \, dy

T1 = SGw * sin(30) * L * \int\limits^h_0 {(h-y)^2} \, dy

T1 = SGw * sin(30) * L * \int\limits^h_0 {(h-y)^2} \, dy

T1 = SGw * sin(30) * L * \int\limits^h_0 {h^2 - 2*h*y + y^2} \, dy

T1 = SGw * sin(30) * L * (h^2*y - h*y^2 + 1/3*y^3)(evaluated between 0 and h)

T1 = SGw * sin(30) * L * (h^2*h - h*h^2 + 1/3*h^3)

T1 = SGw * sin(30) * L * (h^3 - h^3 + 1/3*h^3)

T1 = 1/3 * SGw * sin(30) * L * h^3

To remain stable the equilibrant torque (Teq) must be of larger magnitude than the water pressure torque (T1)

1/4 * b^2 * h * L * SG > 1/3 * SGw * sin(30) * L * h^3

In an equilateral triangle h = b * cos(30)

1/4 * b^3 * cos(30) * L * SG  > 1/3 * SGw * sin(30) * L * b^3 * (cos(30))^3

SG > SGw * 4/3* sin(30) * (cos(30))^2

SG > 1/2 * SGw

For the dam to hold, it should have a specific gravity of at leas half the specific gravity of water.

This is avergae specific gravity, including holes.

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Answer:

Explanation:

<u><em>General Considerations</em></u>

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A glass bottle washing facility uses a well agitated hot water bath at 50°C with an open top that is placed on the ground. The b
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