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Andru [333]
3 years ago
14

The driver of a pickup truck accelerates from rest to a speed of 37 mi/hr over a horizontal distance of 215 ft with constant acc

eleration. The truck is hauling an empty 460-lb trailer with a uniform 72-lb gate hinged at O and held in the slightly tilted position by two pegs, one on each side of the trailer frame at A. Determine the maximum shearing force developed in each of the two pegs during the acceleration.
Physics
1 answer:
ZanzabumX [31]3 years ago
6 0

Answer:

Maximum shearing force developed in each of the two pegs during acceleration is 1830 lbf

Explanation:

First we will find the acceleration of pickup truck.

As, the acceleration is uniform, therefore we can use Newton's third equation of motion:

2as = V_{f}^{2}-V_{i}^{2}

First convert speed into ft/sec

1 mile/hr = 1.47 ft/sec

therefore,

37 mile/hr = 37 x 1.47 ft/sec

37 mile/hr =  54.39 ft/sec

with initial speed 0 ft/sec (starting from rest), using in equation of motion:

a = [(54.39 ft/sec)² - (0 ft/sec)²]/2(215 ft)

a = 6.88 ft/sec²

Now, the total shear force will be given by Newton's second law of motion:

F = ma

F = (460 lbm +72 lbm)(6.88 ft/sec²)

F = 3660 lbf

Now for the max shear force in each of the two pegs we divide total fore by 2:

Force in each peg = F/2 = (3660 lbf)/2

<u>Force in each peg = 1830 lbf</u>

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The energy conservation and trigonometry we can find the results for the questions about the movement of the acrobat are;

     a) The maximum speed is v = 4.89 m / s

     b) The maximum height is h = 1.22 m

The energy conservation is one of the most fundamental principles of physics, stable that if there are no friction forces the mechanistic energy remains constant. Mechanical energy is the sum of the kinetic energy plus the potential energies.

               Em = K + U

Let's write the energy in two points.

Starting point. Highest part of the oscillation

            Em₀ = U = m g h

Final point. Lower part of the movement

            Em_f = K = ½ m v²

Energy is conserved.

            Emo = Em_f  

            m g h = ½ m v²

            v² = 2 gh

Let's use trigonometry to find the height, see attached.

         h = L - L cos θ

         h = L (1- cos θ)

They indicate that the initial angle is tea = 48º and the length is L = 3.7 m, let's calculate.

         h = 3.7 (1- cos 48)

          h = 1.22 m

this  is the maximum height of the movement.

Let's calculate the velocity.  

          v= \sqrt{2 \ 9.8 \ 1.22}  

          v = 4.89 m / s

In conclusion using the conservation of energy and trigonometry we can find the results for the questions about the movement of the acrobat are;

     a) The maximum speed is v = 4.89 m / s

     b) The maximum height is h = 1.22 m

Learn more here: brainly.com/question/13010190

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plasma is a superheated liquid

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The potential of electrons in a circuit can be increased by _____.
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2 years ago
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A vertical spring withk= 245N/m oscillates with an amplitude of 19.2cm when 0.457kg hangs from it. The mass posses through the e
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Answer:

 y = -19.2 sin (23.15t) cm

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Let's look for the terms of this equation the amplitude I

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Angular velocity is

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     w = √ (245 / 0.457

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The φ phase is determined for the initial condition   t = 0 s ,  the velocity is negative v (0) = -vo

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For t = 0

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      φ = sin⁻¹ 1

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Let's build the equation

      y = 19.2 cos (23.15 t + π/ 4)

Let's use the trigonometric ratio π/ 4 = 90º

      Cos (a +90) = cos a cos90 - sin a sin sin 90 = 0 - sin a

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