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monitta
2 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]2 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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______process in sheet metal is used for producing fluid tight joints. A - Hemming B- Seaming C-Beading D-Roll forming
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Answer:

option B

Explanation:

The correct answer is option B

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5 0
2 years ago
A single crystal of a metal that has the FCC crystal structure is oriented such that a tensile stress is applied parallel to the
morpeh [17]

The magnitude of applied stress in the direction of 101 is 12.25 MPA and in the direction of 011, it is not defined.                                          

<u>Explanation</u>:        

<u>Given</u>:

tensile stress is applied parallel to the [100] direction

Shear stress is 0.5 MPA.

<u>To calculate</u>:

The magnitude of applied stress in the direction of [101] and [011].

<u>Formula</u>:

zcr=σ cosФ cosλ

<u>Solution</u>:

For in the direction of 101

cosλ = (1)(1)+(0)(0)+(0)(1)/√(1)(2)

cos λ = 1/√2

The magnitude of stress in the direction of 101 is 12.25 MPA

In the direction of 011

We have an angle between 100 and 011

cosλ = (1)(0)+(0)(-1)+(0)(1)/√(1)(2)

cosλ  = 0

Therefore the magnitude of stress to cause a slip in the direction of 011 is not defined.                                                                                                                                                                      

                                                                                   

                                                                                                                                                   

                                                                                                                                                             

5 0
2 years ago
To unload a bound stack of plywood from a truck, the driver first tilts the bed of the truck and then accelerates from rest. Kno
azamat

Answer:

a) The truck must have an acceleration of at least 3.92 * cos(θ)^2 for the stack to start sliding.

b) a = 2.94 * cos(θ)^2 + 0.32 * L * cos(θ)

Explanation:

The stack of plywood has a certain mass. The weight will depend on that mass.

w = m * g

There will be a normal reaction between the stack and the bed of the truck, this will be:

nr = -m * g * cos(θ)

Being θ the tilt angle of the bed.

The static friction force will be:

ffs = - m * g * cos(θ) * fJK

The dynamic friction force will be:

ffd = - m * g * cos(θ) * fik

These forces would produce accelerations

affs = -g * cos(θ) * fJK

affd = -g * cos(θ) * fik

affs = -9.81 * cos(θ) * 0.4 = -3.92 * cos(θ)

affd = -9.81 * cos(θ) * 0.3 = -2.94 * cos(θ)

These accelerations oppose to movement and must be overcome by another acceleration to move the stack.

The acceleration of the truck is horizontal, the horizontal component of these friction forces is:

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affd = -2.94 * cos(θ)^2

The truck must have an acceleration of at least 3.92 * cos(θ)^2 for the stack to start sliding.

Assuming the bed has a lenght L.

The horizontal movement will be over a distance cos(θ) * L because L is tilted.

Movement under constant acceleration:

X(t) = X0 + V0 * t + 1/2 * a * t^2

In this case X0 = 0, V0 = 0, and a will be the sum of the friction force minus the acceleration of the truck.

If we set the frame of reference with the origin on the initial position of the stack and the positive X axis pointing backwards, the acceleration of the truck will now be negative and the dynamic friction acceleration positive.

L * cos(θ) = 1/2 * (2.94 * cos(θ)^2 - a) * 0.4^2

2.94 * cos(θ)^2 - a = 2 * 0.16 * L * cos(θ)

a = 2.94 * cos(θ)^2 + 0.32 * L * cos(θ)

6 0
2 years ago
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