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gtnhenbr [62]
3 years ago
13

Calculate the reactions at 4 ends (supports) of this bookshelf. Assume that the weight of each book is approximately 1 lb. The w

idth of shelf is 10inches and its length is 40 inches. (Hint: Model the shelf as a beam with distributed load).

Engineering
1 answer:
hoa [83]3 years ago
8 0

Answer:

R = 4.75 lb  (↑)

Explanation:

Number of books = n = 19

Weight of each book = W = 1 lb

Length of the bookshelf = L = 40 inches

We can get the value of the distributed load as follows

q = n*W/L = 19*1 lb/ 40 inches = 0.475 lb/in

then the reactions at 4 ends (supports) of the bookshelf are

R = (q/2)/2 = 4.75 lb

We can see the bookshelf in the pic.

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A three-point flexure test is conducted on a cylindrical specimen of aluminum oxide. The specimen radius is 5.0 mm and the dista
kondaur [170]

Answer:

Detailed solution is given in the attached diagram

7 0
3 years ago
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.

6 0
2 years ago
Prefix version of 6600 volts​
GenaCL600 [577]

Answer:

6.6 kilo volts = 6.6 k volts

Explanation:

A prefix is a word, number or a letter that is added before another word. In physics we have different prefixes for the exponential powers of 10, that are placed before units in place of those powers. Some examples are:

deci (d)   ------  10⁻¹

centi (c)   ------  10⁻²

milli (m)   ------   10⁻³

kilo (k)     ------   10³

mega (M) -----   10⁶

giga (G)   ------   10⁹

We have:

6600 volts

converting to exponential form:

=> 6.6 x 10³ volts

Thus, we know that the prefix of kilo (k) is used for 10³.

Hence,

=> <u>6.6 kilo volts = 6.6 k volts</u>

7 0
3 years ago
The pressure of a gas in a rigid container is 125kpa at 300k, what we be the new pressure if the temperature increases to 900k​
kipiarov [429]

Answer:

375 KPa

Explanation:

From the question given above, the following data were obtained:

Initial pressure (P₁) = 125 KPa

Initial temperature (T₁) = 300 K

Final temperature (T₂) = 900 K

Final pressure (P₂) =?

The new (i.e final) pressure of the gas can be obtained as follow:

P₁/T₁ = P₂/T₂

125 / 300 = P₂ / 900

Cross multiply

300 × P₂ = 125 × 900

300 × P₂ = 112500

Divide both side by 300

P₂ = 112500 / 300

P₂ = 375 KPa

Thus, the new pressure of the gas is 375 KPa

7 0
2 years ago
Determine the velocity of the 13-kgkg block BB in 4 ss . Express your answer to three significant figures and include the approp
Anvisha [2.4K]

Answer:

The question has some details missing : The 35-kg block A is released from rest. Determine the velocity of the 13-kgkg block BB in 4 ss . Express your answer to three significant figures and include the appropriate units. Enter positive value if the velocity is upward and negative value if the velocity is downward.

Explanation:

The detailed steps and appropriate calculation is as shown in the attached file.

6 0
3 years ago
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