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vesna_86 [32]
3 years ago
5

Calculate the factor of safety against bending failure of a gear with a given tooth bending stress of 117MPa and a corrected ben

ding strength of 354MPa.
Engineering
1 answer:
Deffense [45]3 years ago
3 0

Answer:

3

Explanation:

Bending stress = 117 MPa

Yield stress = 354 MPa

Factor of safety is defined as the ratio of yield stress to the allowable stress (tensile, compressive or bending) of a given material

FS= \frac{\text{Yield stress}}{\text{Allowable stress}}\\\Rightarrow FS=\frac{354}{117}\\\Rightarrow FS=3.025

So, factor of safety is almost equal to 3.

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A tool chest has 650 N weight that acts through the midpoint of the chest. The chest is supported by feet at A and rollers at B.
erica [24]

Answer:

the value of horizontal force P is 170.625 N

the value of horizontal force at P = 227.5 N is that the block moves to right and this motion is due to sliding.

Explanation:

The first diagram attached below shows the free body diagram of the tool chest when it is sliding.

Let start out by calculating the friction force

F_f= \mu N_2

where :

F_f = friction force

\mu = coefficient of friction

N_2 = normal friction

Given that:

\mu = 0.3

F_f = 0.3 N_2

Using the equation of equilibrium along horizontal direction.

\sum f_x = 0

P - F_f = 0

P = 0.3 N_2   ----- Equation (1)

To determine the moment about point B ; we have the expression

\sum M_B  = 0

0 = N_2*70-W*35-P*100

where;

P = horizontal force

N_2 = normal force at support A

W = self- weight of tool chest

Replacing W = 650 N

0 = N_2*70-650*35-100*P

P = \frac{70 N_2-22750}{100} ----- equation (2)

Replacing  \frac{70 N_2-22750}{100}  for P in equation (1)

\frac{70N_2 -22750}{100} =0.3 N_2

N_2 = \frac{22750}{40}N_2 = 568.75 \ N

Plugging the value of N_2 = 568.75 \ N in equation (2)

P = \frac{70(568.75)-22750}{100} \\ \\ P = \frac{39812.5-22750}{100}  \\ \\ P = \frac{17062.5}{100}

P =170.625 N

Thus; the value of horizontal force P is 170.625 N

b)  From the second diagram attached the free body diagram; the free body diagram of the tool chest when it is tipping about point A is also shown below:

Taking the moments about point A:

\sum M_A = 0

-(P × 100)+ (W×35) = 0

P = \frac{W*35}{100}

Replacing 650 N  for W

P = \frac{650*35}{100}

P = 227.5 N

Thus; the value of horizontal force P, when the tool chest tipping about point A is 227.5 N

We conclude that the motion will be impending for the lowest value when P = 170.625 N and when P= 227.5 N

However; the value of horizontal force at P = 227.5 N is that the block moves to right and this motion is due to sliding.

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Answer:

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How many broadcast or vlan is in this switchs and router ? and why? ​
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Problem: Big Broadcast Domains

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The left side of Figure 4-1 depicts a small network in which PC 2 and PC 4 attempt transmissions at the same time. The frames propagate away from the computers, eventually colliding with each other somewhere in between the two nodes as shown on the right. The increased voltage and power then propagate away from the scene of the collision. Note that the collision does not continue past the switches on either end. These are the boundaries of the collision domain. This is one of the primary reasons for switches replacing hubs. Hubs (and access points) simply do not scale well as network traffic increases.

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