College

Consider a cantilever beam with a uniformly distributed load between 90 and 105 pounds per linear foot.

a. What is the probability that a beam load exceeds 99 pounds per linear foot?

b. What is the probability that a beam load is less than 92 pounds per linear foot?

c. Find a value \( L \) such that the probability that the beam load exceeds \( L \) is 0.4.

Answer :

Answer:

a) 0.4

b) 0.133

c) [tex]L \geq 99[/tex]

Step-by-step explanation:

We are given the following information in the question:

The load is said to be uniformly distributed over that part of the beam between 90 and 105 pounds per linear foot.

a = 90 and b = 105

Thus, the probability distribution function is given by

[tex]f(x) = \displaystyle\frac{1}{b-a} = \frac{1}{105-90} = \frac{1}{15},\\\\90 \leq x \leq 105[/tex]

a) P( beam load exceeds 99 pounds per linear​ foot)

P( x > 99)

[tex]=\displaystyle\int_{99}^{105} f(x) dx\\\\=\displaystyle\int_{99}^{105} \frac{1}{15} dx\\\\=\frac{1}{15}[x]_{99}^{105} = \frac{1}{15}(105-99) = 0.4[/tex]

b) P( beam load less than 92 pounds per linear​ foot)

P( x < 92)

[tex]=\displaystyle\int_{90}^{92} f(x) dx\\\\=\displaystyle\int_{90}^{92} \frac{1}{15} dx\\\\=\frac{1}{15}[x]_{90}^{92} = \frac{1}{15}(92-90) = 0.133[/tex]

c) We have to find L such that

[tex]\displaystyle\int_{L}^{105} f(x) dx\\\\=\displaystyle\int_{L}^{105} \frac{1}{15} dx\\\\=\frac{1}{15}[x]_{L}^{105} = \frac{1}{15}(105-L) = 0.4\\\\\Rightarrow L = 99[/tex]

The beam load should be greater than or equal to 99 such that the probability that the beam load exceeds L is 0.4.

Final answer:

The probability that the load exceeds 99 pounds per linear foot is 40%. There's a 13.33% chance that the load is less than 92 pounds per linear foot. The value L such that there is a 40% chance the load exceeds it is 99 pounds per linear foot.

Explanation:

To solve problems related to uniformly distributed loads on beams, we use the principles of probabilities. Since the load is uniformly distributed between 90 and 105 pounds per linear foot, each value within this range is equally likely.

a. Probability of exceeding 99 pounds per linear foot

The range of distribution is 105 - 90 = 15 pounds per linear foot. To find the probability that the load exceeds 99 pounds per linear foot, we calculate:

Probability = (Upper limit - Desired value) / Total range

Probability = (105 - 99) / 15 = 6 / 15 = 0.4

Thus, there's a 40% chance that the load exceeds 99 pounds per linear foot.

b. Probability of being less than 92 pounds per linear foot

Similarly, to find the probability that the load is less than 92 pounds:

Probability = (Desired value - Lower limit) / Total range

Probability = (92 - 90) / 15 = 2 / 15 ≈ 0.1333

Therefore, there's a roughly 13.33% chance that the load is less than 92 pounds per linear foot.

c. Finding value L for a 0.4 probability of exceeding the load

To find the value L such that the probability of exceeding it is 0.4, we work backward:

0.4 = (105 - L) / 15

L = 105 - (0.4 * 15)

L = 105 - 6

L = 99 pounds per linear foot

So, the value of L for which there is a 40% chance that the load exceeds it is 99 pounds per linear foot.