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poizon [28]
3 years ago
10

The alignment readings for the front of a vehicle are shown above. Camber and toe are within specification, caster is not. Techn

ician A says that with the current settings, the left front tire tread may wear on the inside edge. Technician B says that with the current settings, the vehicle may pull to the left. Who is right? A. A only B. B only C. Both A and B D. Neither A nor B
Engineering
1 answer:
dlinn [17]3 years ago
4 0

Answer:

B. B only

Given Information:

1. Camber and toe are within specification

2. Caster is not within specification

Technician A says that with the current settings, the left front tire tread may wear on the inside edge.

Technician B says that with the current settings, the vehicle may pull to the left

Explanation:

Lets discuss the effects of Camber, toe and caster misalignment

Effects of Camber and Toe misalignment:

Camber is the inward or outward tilt of the fron tires and is used to distribute load across the tread. Any misalignment causes uneven loading on the tires which results in tire wear on one edge.

The most common cause of tire wear on the inside edge is due to the camber misalignment which results in premature tire wear.

Another reason is of tire wear is vehicle’s toe. A slight misalignment of the toe reduces the life of the tire.

Since it is given that camber settings and toe settings are within specification therefore, tire tread wear on the inside edge cannot happen if camber and toe are within specification.

Technician A cannot be right.

Effects of Caster misalignment:

Whenever there is a misalignment of the castor then the vehicle will not be able to go in straight line rather it will pull to either left or right side. Caster misalignment also causes heavy or light steering depending upon the positive or negative misalignment of caster.

Since it is given that caster settings are not within specification therefore, the vehicle may pull to the left due to the caster misalignment.

Technician B must be right.

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A fatigue test is performed on 69 rotating specimens made of 5160H steel. The measured number of cycles to failure (L in kcycles
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Answer:

(a) Mean = 122.9, σ = 30.071

(b) No. of failed specimens at less than 115k cycles are 27.

(c) μ = 39.07

Explanation:

We are given:

L  60  70  80  90  100  110  120  130  140  150  160  170  180  190  200  210

f    2     1    3     5     8     12     6     10     8     5     2      3      2      1       0      1

(a) First we need to calculate the mean and standard deviation. The formula for calculating mean is:

Mean = ∑fx/∑f

And for standard deviation we have:

S.D. = √Var

Var = ∑fx²/∑f - (Mean)²

∑fx = (2*60) + (1*70) + (3*80) + (5*90) + (8*100) + (12*110) + (6*120) + (10*130) + (8*140) + (5*150) + (2*160) + (3*170) + (2*180) + (1*190) + (0*200) + (1*210)

         = 120 + 70 + 240 + 450 + 800 + 1320 + 720 + 1300 + 1120 + 750 + 320 + 510 + 360 + 190 + 0 + 210

∑fx = 8480

Mean = ∑fx/∑f

          = 8480/69

Mean = 122.9  

∑fx² = (2*60²) + (1*70²) + (3*80²) + (5*90²) + (8*100²) + (12*110²) + (6*120²) + (10*130²) + (8*140²) + (5*150²) + (2*160²) + (3*170²) + (2*180²) + (1*190²) + (0*200²) + (1*210²)

   =7200+4900+19200+40500+80000+145200+86400+169000+156800+112500+51200+86700+64800+36100+0+44100

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Var = ∑fx²/∑f - (Mean)²

     = 1104600/69 - (122.9)²

     = 16008.69565 - 15104.41

Var = 904.2856

S.D = √Var

σ = √904.2856

σ = 30.071

(b) Let X be the number of failed specimen.

We will use the z-score to calculate the probability. The formula for z-score is:

z = (X-μ)/σ

P(X<115) = P(z<(115-122.9)/30.071)

              = P(z<-0.26)

Using the normal distribution probability table, we can compute the value of  P(z<-0.26).

P(X<115) = 0.3974

So, no. of failed specimens at less than 115k cycles are: 0.3974*69 = 27 specimens

(c) σ = 30.071

P(x<115) = 0.99

P(z<(115-μ)/30.071) = 0.99

From the normal distribution table we find that 0.99 lies between the z values 2.52 and 2.33. Hence, we get 2.525 as the z-value at which the probability is 0.99.

z = (x-μ)/σ

2.525 = (115 - μ)/30.071

75.93 = 115 - μ

μ = 115 - 75.93

μ = 39.07

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