The sum of the magnitudes of two forces acting at a point is 18 and th...
Problem: Given the sum of the magnitudes of two forces acting at a point is 18 and the magnitude of their resultant is 12. If the resultant is at 90 degrees with the force of smaller magnitude, what are the magnitudes of forces?
Solution:
Let the magnitudes of two forces be x and y. We know that the sum of their magnitudes is 18, so:
x + y = 18 ... (1)
We also know that the magnitude of their resultant is 12 and the angle between the resultant and the force of smaller magnitude is 90 degrees. So, we can use Pythagoras theorem to find the value of the other side of the right triangle formed by the two forces and their resultant:
x^2 + y^2 = 12^2 ... (2)
Now, we can solve the two equations (1) and (2) simultaneously to find the values of x and y:
From equation (1), we have:
y = 18 - x
Substituting this in equation (2), we get:
x^2 + (18 - x)^2 = 144
Simplifying this equation, we get:
2x^2 - 36x + 36 = 0
Dividing both sides by 2, we get:
x^2 - 18x + 18 = 0
Using the quadratic formula, we get:
x = 9 + 3√5 or x = 9 - 3√5
Since the force of smaller magnitude cannot be negative, we take:
x = 9 - 3√5
Substituting this value in equation (1), we get:
y = 18 - x = 9 + 3√5
Therefore, the magnitudes of the two forces are:
x = 9 - 3√5
y = 9 + 3√5
Conclusion: The magnitudes of the two forces are 9 - 3√5 and 9 + 3√5.
The sum of the magnitudes of two forces acting at a point is 18 and th...
It's given that resultant makes right angle with A(smaller force).hence,B must be hypotenuse
So B^2=R^2+A^2
=>R^2 =B^2-A^2= 12^2= 144 (1)
It is also given that A+B = 18
B= 18-A (2)
Substitute (2) in (1) (18-A)^2-A^2= 144
324 - 36A +A^2 -A^2=144
36A = 324 –144 = 180
A =180/36 = 5
Hence, we get B = 18 - A = 18-5= 13
Therefore, the magnitude of the two vectors are 5 and 13
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