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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 acceleration. 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.

User Oh My Dog
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2 Answers

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Final answer:

To find the maximum shearing force in the pegs during the truck's acceleration, Newton's second law is applied, and the weight and acceleration of the gate are considered. The final shear force for each peg is 7.7021 lbs.

Step-by-step explanation:

To find the maximum shearing force in each peg during the acceleration of the pickup truck with the trailer, we need to apply Newton's second law.

First, let's convert the speed from miles per hour to feet per second (1 mi/hr = 1.46667 ft/s):

37 mi/hr × 1.46667 ft/s/mi/hr = 54.2667 ft/s

Using the kinematic equation v^2 = u^2 + 2as (where v is final velocity, u is initial velocity, s is distance, and a is acceleration), we can solve for acceleration (a) since the truck starts from rest (u = 0):

54.2667^2 = 0 + 2 × a × 215

a = × 54.2667^2 / (2 × 215) = 6.88776 ft/s^2

The force affecting the gate pegs comes from the horizontal component of the gravitational force and the force due to acceleration. Assume that the mass of the gate is 72 lbs; the gravitational force is:

F_gravity = m × g = 72 lbs (Since g in lbs already accounts for Earth's gravitational acceleration)

To find the force due to acceleration, we use F = m × a:

F_acceleration = 72 lbs × 6.88776 ft/s^2

However, since lbs already include g, we convert lbs to mass in slugs:

m = 72 lbs / 32.2 ft/s^2 = 2.23602 slugs

F_acceleration = 2.23602 slugs × 6.88776 ft/s^2 = 15.4042 lbs

The total force on each peg is thus the shearing force due to the acceleration, but as there are two pegs, we divide the total force by 2, assuming the load is distributed evenly:

Shear force per peg = F_acceleration / 2 = 15.4042 lbs / 2 = 7.7021 lbs (rounded to four significant figures)

User Shaoyihe
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3 votes

Answer:

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

Step-by-step 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

Force in each peg = 1830 lbf

User Lokimidgard
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