The maximum load m of a beam varies directly as the breadth b and the square of the depth d and inversely as the length l.If a beam 3.6m long,1.2m wide,and 2.4m deep can safely bear a load up to 909kg,find the maximum safe load for a beam of the same material which is 3m long,6m wide,and 1.8m deep
If the load is called z, then
z = kbd^2/L
Thus, zL/bd^2 = k is constant
So you want to find z such that
(z*3)/(6*1.8^2) = (909*3.6)/(1.2*2.4^2)
z = 3067.88 kg
Step 1: Write down the given information:
Length of the first beam (l1) = 3.6 m
Width of the first beam (b1) = 1.2 m
Depth of the first beam (d1) = 2.4 m
Maximum load for the first beam (m1) = 909 kg
Length of the second beam (l2) = 3 m
Width of the second beam (b2) = 6 m
Depth of the second beam (d2) = 1.8 m
Step 2: Use the direct proportionality equation to relate the maximum load (m) with the breadth (b) and depth (d):
m ∝ b * d^2
Step 3: Use the inverse proportionality equation to relate the maximum load (m) with the length (l):
m ∝ 1 / l
Step 4: Write the equation combining the direct and inverse proportionality:
m = k * (b * d^2) / l
where k is the constant of proportionality.
Step 5: Use the given information to find the constant of proportionality (k). Substitute the values of the first beam into the equation:
909 = k * (1.2 * 2.4^2) / 3.6
Step 6: Simplify and solve for k:
k = (909 * 3.6) / (1.2 * 2.4^2) = 757.5
Step 7: Now, we can use the value of k to find the maximum safe load for the second beam. Substitute the values of the second beam into the equation:
m2 = (757.5 * (6 * 1.8^2)) / 3 = 7575 kg
Therefore, the maximum safe load for a beam of the same material, which is 3m long, 6m wide, and 1.8m deep, is 7575 kg.
To find the maximum safe load for the second beam, we can use the concept of direct and inverse variation provided in the question. We are given the measurements of two beams and their corresponding maximum safe load values, and we need to find the maximum safe load for the second beam.
Let's denote the maximum safe load of the first beam as M1, the breadth of the first beam as b1, the depth of the first beam as d1, and the length of the first beam as l1. Similarly, let's denote the maximum safe load of the second beam as M2, the breadth of the second beam as b2, the depth of the second beam as d2, and the length of the second beam as l2.
According to the given information, we have the following relationships:
1. M1 is directly proportional to (b1 * d1^2)
2. M1 is inversely proportional to l1
Using these relationships, we can form an equation:
M1 = k * (b1 * d1^2) / l1
where k is the constant of proportionality.
We can rearrange this equation to solve for k:
k = M1 * l1 / (b1 * d1^2)
Now, we can substitute the given values for the first beam into the equation to find the value of k:
k = 909 kg * 3.6 m / (1.2 m * (2.4 m)^2)
Simplifying this expression:
k = 0.126875 kg/m^4
Now, we can use this value of k to determine the maximum safe load for the second beam.
M2 = k * (b2 * d2^2) / l2
Substituting the given values for the second beam and the previously calculated value of k:
M2 = (0.126875 kg/m^4) * (6 m * (1.8 m)^2) / 3 m
Simplifying this expression:
M2 = 0.126875 kg/m^4 * 19.512 kg/m
M2 ≈ 2.48 kg
Therefore, the maximum safe load for a beam of the same material which is 3m long, 6m wide, and 1.8m deep is approximately 2.48 kg.