A string of length l carries a small pulley C from which is suspended a load W. The string hangs between two vertical walls which are at distance d apart.The end A is higher than end B by height h.find position of equilibrium defined by angle X. X is angle between AC and AB.
d=l/2
h=l/4
To solve this problem, let's first visualize the scenario. We have a string of length l hanging between two vertical walls, with a small pulley C attached to it. Hanging from the pulley is a load W. The end A of the string is higher than the end B by a height h, and the distance between the walls is d, which is half the length of the string (d = l/2).
Now, let's find the position of equilibrium defined by the angle X between AC and AB.
To start, draw a diagram with the string hanging between the two walls. Draw a line segment AB vertically, where A is higher than B by a height h. From point A, draw a line segment AC that is perpendicular to AB, forming a right triangle ABC.
We can use the properties of right triangles to help us solve this problem. As triangle ABC is a right triangle, we can use trigonometric ratios.
Since we know the values of d and h, we can find the length of the hypotenuse AC using the Pythagorean theorem:
AC^2 = AB^2 + BC^2
Since AB is the height h and BC is the distance d, we can rewrite the equation as:
AC^2 = h^2 + d^2
Now let's substitute the values of d and h into the equation:
AC^2 = (l/4)^2 + (l/2)^2
Simplifying further:
AC^2 = l^2/16 + l^2/4
AC^2 = (l^2/16) + (4l^2/16)
AC^2 = (l^2 + 4l^2)/16
AC^2 = 5l^2/16
Now take the square root of both sides to find the length of AC:
AC = sqrt(5l^2/16)
AC = (sqrt(5) * l)/4
Now, to find the angle X, we can use the trigonometric ratio:
tan(X) = opp/adj
tan(X) = BC/AB
tan(X) = d/h
Substituting the values of d and h:
tan(X) = (l/2) / (l/4)
tan(X) = 2/1
Take the arctan to find the angle X:
X = arctan(2)
So, the position of equilibrium defined by the angle X is arctan(2) radians.
To find the position of equilibrium defined by angle X, we can use basic trigonometry.
First, let's visualize the situation. We have a string of length l hanging between two vertical walls, with end A higher than end B by height h. Pulley C is suspended from the string, and there is a load W attached to it.
By the given information, we know that the distance between the two walls is d = l/2. We also know that the height difference between ends A and B is h = l/4.
To find the position of equilibrium defined by angle X, we need to calculate the length of the string on each side of the pulley. Let's call the length of string on side CB as 'd1' and the length of string on side CA as 'd2'.
Using the Pythagorean theorem, we can relate the distances d1, d2, and l as follows:
d1^2 + h^2 = l^2 ---(1)
d2^2 + h^2 = l^2 ---(2)
Since d1 + d2 = d (the distance between the walls), we can rewrite d1 and d2 in terms of l and d:
d1 = (d - d2)
d2 = (d - d1)
Substituting these values into equations (1) and (2), we get:
(d - d2)^2 + h^2 = l^2
(d - d1)^2 + h^2 = l^2
Simplifying these equations:
(d - d2)^2 - l^2 = -h^2
(d - d1)^2 - l^2 = -h^2
Expanding the equations:
d^2 - 2dd2 + d2^2 - l^2 = -h^2
d^2 - 2dd1 + d1^2 - l^2 = -h^2
Rearranging the equations, we have:
d2^2 - 2dd2 + l^2 = h^2
d1^2 - 2dd1 + l^2 = h^2
Since d1 + d2 = d, we can rewrite these equations as:
d1^2 - 2d(d - d1) + l^2 = h^2
d2^2 - 2d(d - d2) + l^2 = h^2
Expanding and simplifying:
d1^2 - 2d^2 + 2dd1 + l^2 = h^2
d2^2 - 2d^2 + 2dd2 + l^2 = h^2
Simplifying further:
d1^2 + l^2 - 2d^2 + 2dd1 = h^2
d2^2 + l^2 - 2d^2 + 2dd2 = h^2
Now we have two equations with two unknowns (d1 and d2). We can solve these equations simultaneously to find their values. Once we have the values of d1 and d2, we can calculate the tangent of angle X using the equation:
tan(X) = h / (d1 - d2)
Finally, we have the position of equilibrium defined by angle X.