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kobusy [5.1K]
1 year ago
7

the uniform bar (mass m and length l ) is leaning against a wall. the coefficient of static friction between all surfaces and th

e bar ends is 0.30. determine the value for p which causes impending slip if p acts to (a) the right and (b) the left. (points 20)
Engineering
1 answer:
zaharov [31]1 year ago
8 0

The coefficient of static friction is the ratio of the maximum static friction force (F) between the surfaces in contact before movement commences to the normal (N) force.

<h3>What is coefficient of static friction formula?</h3>
  • The friction coefficient is the ratio of the normal force pushing two surfaces together to the frictional force preventing motion between them.
  • Typically, it is represented by the Greek letter mu (). In terms of math, is equal to F/N, where F stands for frictional force and N for normal force.
  • The ratio of the greatest static friction force (F) between the surfaces in contact before movement starts to the normal force (N) is known as the coefficient of static friction.
  • A body and a surface have static and kinetic friction coefficients of 0.75 and 0.5, respectively. The body is forced to slide with a constant acceleration that is equal to A. g4 by applying a force.

Find the attachment answer.

Learn more about coefficient of static friction refer to :

brainly.com/question/25050131

#SPJ4

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Determine the time required for a car to travel 1 km along a road if the car starts from rest, reaches a maximum speed at some i
lbvjy [14]

Total time taken by car to travel 1 km distance is 48.2 seconds

Explanation:

Given data-

Total distance- 1 km= 1000m

Car starts from rest- Hence the initial velocity (u)= 0 m/s

Then, car accelerates at 1.5 m/s ²

Let us suppose with these acceleration car reaches the max speed of V

Car then decelerates at rate of 2 m/s ²

Finally, car comes to rest

We need to consider the question in two parts

Part 1= when car starts from rest and reaches the value V in the time tₐ at the distance s₁.

Part 2= When car starts from V and finally stops at 1 km mark in the time tₙ in distance 1000-s₁.

For the part 1

We know the formula  

v= u + a*tₐ

where v= final velocity

u- initial velocity

a= acceleration

tₐ= time period

At the starting u= 0  

Hence the equation reduces to V=0+1.5tₐ

Or V= 1.5tₐ                             Eq 1

We also know that s= u*tₐ+ ¹/₂*a*tₐ²

Where s₁= distance covered  (other symbols same meaning)

Since u=0   (u*tₐ=0)

s₁ = ¹/₂*1.5*tₐ²

s₁= ½* 1.5*tₐ²                    -------------Eq 2

Now considering Part 2

Here the case is deceleration hence the equation would change (symbols same)

v= V-a*tₙ     final velocity(v)=0 (car stops finally) & initial velocity (part 2)= V

V= a*tₙ

V=2*tₙ                                 -----------Eq 3

Similarly  

1000-s₁= V* tₙ+¹/₂*(-2) *(tₙ)²

1000-s₁= V* tₙ-tₙ²                      -------Eq 4

Comparing Eq 1 and Eq 3

V= 1.5tₐ  and V=2tₙ                                                              

1.5tₐ = 2 tₙ

tₐ=1.33 tₙ  

Using the above value of tₐ in Eq 1

V= 1.5 tₐ and tₐ= 1.33 tₙ

V= 2tₙ

Similarly from Eq 2 and putting the value of tₙ

s₁= ½*1.5*tₐ²      

s₁=  1.33*(tₙ)²

Substituting the above values in equation 4

1000-s₁= V* tₙ-tₙ²                      

1000- 1.33(tₙ)²=2*(tₙ)*(tₙ)- tₙ²

1000=2.33 (tₙ)²

tₙ=      1000/2.333    

tₙ= 20.7 sec

Similarly putting the value of tₙ in tₐ= 1.33 tₙ

tₙ= 27.5 sec

Hence total time is tₐ +tₙ  

T= 20.7+27.5= 48.2 sec                

8 0
3 years ago
Steel balls 10 mm in diameter are annealed by heating to 1150 k and then slowly cooling to 450 k in an air environment for which
Darina [25.2K]

Answer:

The answer is below

Explanation:

Given that:

ρ = 800 kg/m3, c = 600 J/kg-K,  k = 40 W/m-K, Initial temperature = Ti = 1150,

Environment temperature = T = 450 K, Final temperature = T∞ = 325 K

Diameter = 10 mm = 0.01 m, A = 6

The estimated time for cooling process (t) is given as:

t=\frac{\rho dc}{hA} ln\frac{T_i-T_\infty}{T-T_infty}=\frac{7800*0.01*600}{25*6} ln\frac{1150-325}{450-325}\\  \\t=589\ s\\t=0.1635\ h

The estimated cooling time is 589 s

4 0
3 years ago
1. Two aluminium strips and a steel strip are to be bonded together to form a composite bar. The modulus of elasticity of steel
yarga [219]

Answer:

1.933 KN-M

Explanation:

<u>Determine the largest permissible bending moment when the composite bar is bent  horizontally </u>

Given data :

modulus of elasticity of steel = 200 GPa

modulus of elasticity of aluminum = 75 GPa

Allowable stress for steel = 220 MPa

Allowable stress for Aluminum = 100 MPa

a = 10 mm

<em>First step </em>

determine moment of resistance when steel reaches its max permissible stress

<em>next </em>: determine moment of resistance when Aluminum reaches its max permissible stress

Finally Largest permissible bending moment of the composite Bar = 1.933 KN-M

<em>attached below is a detailed solution </em>

5 0
3 years ago
How much will it cost to train the entire company to use a recycling program if the training includes paper handouts? (Remember,
bezimeni [28]

Answer:

With the current budget of $1,000, it is possible to train the company and buy 3 recycling carts given that recycling pickup within the area is free

Explanation:

The office waste management budget that is not being spent = $1,000

The given expense parameters are;

The training cost per hour per employee = $12.00

The number of people in the entire company = 45

The cost of preparing the training material per hour = $20.00

The number of people to create the training material = 3 people

The time it will take each person in creating the training material = 3 hours

The cost of recycling cart = $80.00 per cart

The cost of paper handouts per employee = $0.05

The cost of materials include;

The total cost for the training = $12.00 × 45 = $540

The cost for the handout = $0.05 × 45 = $2.25

The cost of preparing the materials = $20.00 × 3 × 3 = $180.00

The total costs of the training = $540 + $4 + $180.00 = $724

The amount available to buy cart = $1,000 - $724 = $276

Therefore;

The amount available to buy cart = $276

The number of carts that can be bought = 276/80 = 3.45 carts

Therefore, we round down to get;

The number of carts that can be bought = 3 carts

8 0
3 years ago
What advantage might there be to having the encoder located on the motor side of the gearhead instead of at the output shaft of
pickupchik [31]

The most accurate answer to that process is definitely precision. The Rotary encoder is an electro-mechanical device that converts the angular position or motion of a shaft or axle to analog or digital output signals. The efficiency of these devices is subject to the position and angle of the axis in front of the encoder.

Most cars use reduction systems in their gearboxes that convert a certain signal input into an output. Mechanically for example, a 20: 1 reduction box already infers that if there is a revolution in the input at the output there are 20. That same transferred to the encoder pulses would imply greater precision.

For example a decoder with 50 holes would have to read 1000 pulses (50 * 20) which is basically a degree of accuracy of 0.36 degrees. In this way it is possible to conclude that if the assembly of the encoder is carried out next to the motor and not at the output, it can be provided with greater precision at the time of reading.

7 0
4 years ago
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