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Vikentia [17]
3 years ago
7

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

wing that the coefficients of friction between the bottom sheet of plywood and the bed are fJK = 0.40 and fik = 0.30, determine (a) the smallest acceleration of the truck which will cause the stack of plywood to slide, (b) the acceleration of the truck which causes corner A of the stack of plywood to reach the end of the bed in 0.4 s.
Engineering
1 answer:
azamat3 years ago
6 0

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:

affs = -3.92 * cos(θ)^2

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)

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8 0
3 years ago
Consider casting a concrete driveway 40 feet long, 12 feet wide and 6 in. thick. The proportions of the mix by weight are given
Akimi4 [234]

Answer:

Weight of cement = 10968 lb

Weight of sand = 18105.9 lb

Weight of gravel = 28203.55 lb

Weight of water = 5484 lb

Explanation:

Given:

Entrained air = 7.5%

Length, L = 40 ft

Width,w = 12 ft

thickness,b= 6 inch, convert to ft = 6/12 = 0.5 ft

Specific gravity of sand = 2.60

Specific gravity of gravel = 2.70

The volume will be:

40 * 12 * 0.5 = 240 ft³

We need to find the dry volume of concrete.

Dry volume = wet volume * 1.54 (concrete)

Dry volume will be = 240 * 1.54 = 360ft³

Due to the 7% entarained air content, the required volume will be:

V = 360 * (1 - 0.07)

V = 334.8 ft³

At a ratio of 1:2:3 for cement, sand, and gravel respectively, we have:

Total of ratio = 1+2+3 = 6

Their respective volume will be =

Volume of cement = \frac{1}{6}*334.8 = 55.8 ft^3

Volume of sand = \frac{2}{6}*334.8 = 111.6 ft^3

Volume of gravel = \frac{3}{6}*334.8 = 167.4 ft^3

To find the pounds needed the driveway, we have:

Weight = volume *specific gravity * density of water

Specific gravity of cement = 3.15

Weight of cement =

55.8 * 3.15 * 62.4 = 10968 pounds

Weight of sand =

111.6 * 2.60 * 62.4 = 18105.9 lb

Weight of gravel =

167.4 * 2.7 * 62.4 = 28203.55 lb

Given water to cement ratio of 0.50

Weight of water = 0.5 of weight of cement

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4 0
3 years ago
Which term represents an object that has a round or oval base and is connected at every point by lines at a corresponding point
raketka [301]

Answer:

it is a polyhedron

Explanation:

if I am wrong I am sorry

8 0
3 years ago
Read 2 more answers
Find the mechanical average of a wheel axle System of the wheel has a radius of 1.5 feet in the accident has a radius of 6 inche
kvasek [131]

Answer:

Mechanical average of a wheel = 3

Explanation:

Given:

Radius of wheel = 1.5 ft = 1.5 x 12 = 18 inches

Radius of axle = 6 inches

Find:

Mechanical average of a wheel

Computation:

Mechanical average of a wheel = Radius of wheel / Radius of axle

Mechanical average of a wheel = 18 / 6

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4 0
3 years ago
A multipurpose transformer has a secondary coil with several points at which a voltage can be extracted, giving outputs of 5.60,
Rudik [331]

Answer:

for 5.6V 9 turns, for 12.0V 19 turns, for 480V 755 turns

Explanation:

Vp/Vs= Np/Ns

Vp: Primary voltage

Vs: Secondary Voltage

Np: number of turns on primary side

Ns: number of turns on secondary side

for output 5.6V

140/5.6= 220/Ns

Ns= 8.8 or 9 Turns

for output 12.0V

140/12= 220/Ns

Ns= 18.9 or 19 turns

for output 480V

140/480= 220/Ns

Ns= 754.3 or 755 turns

4 0
3 years ago
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