A rope that is 6 m long is fixed between two points that are separated by 5 m and are both horizontal to each other. A 10 kg weight is attached to the rope at 1 m from the left end. This would cause the rope to form the hypotenuse of two right triangles. The first hypotenuse would have a length of one, and the second would have a length of five. The two hypotenuse is would touch each other at where the weight Was attached. The other end of the hypotenuses Would be the end of the rope. Calculate the angle that both hypotenuse is make with respect to horizontal.
To solve this problem, we need to determine the angles that the two segments of the rope make with the horizontal. Let's denote the points as follows:
Given:
We need to find the angles that the segments ( AC ) and ( CB ) make with the horizontal.
Since the weight is attached 1 meter from ( A ), the length of ( AC ) is 1 meter. The remaining length of the rope, ( CB ), is ( 6 - 1 = 5 ) meters.
Let's denote the vertical distances from ( A ) to ( C ) as ( h_1 ) and from ( C ) to ( B ) as ( h_2 ).
For ( AC ): [ AC = \sqrt{(1)^2 + (h_1)^2} ] Since ( AC = 1 ) meter: [ 1 = \sqrt{(1)^2 + (h_1)^2} ] [ 1 = \sqrt{1 + (h_1)^2} ] [ 1^2 = 1 + (h_1)^2 ] [ 1 = 1 + (h_1)^2 ] [ (h_1)^2 = 0 ] [ h_1 = 0 ]
For ( CB ): [ CB = \sqrt{(4)^2 + (h_2)^2} ] Since ( CB = 5 ) meters: [ 5 = \sqrt{(4)^2 + (h_2)^2} ] [ 5 = \sqrt{16 + (h_2)^2} ] [ 5^2 = 16 + (h_2)^2 ] [ 25 = 16 + (h_2)^2 ] [ (h_2)^2 = 9 ] [ h_2 = 3 ]
For ( AC ): Since ( h_1 = 0 ), the angle ( \theta_1 ) that ( AC ) makes with the horizontal is: [ \theta_1 = \arctan\left(\frac{h_1}{1}\right) = \arctan(0) = 0^\circ ]
For ( CB ): [ \theta_2 = \arctan\left(\frac{h_2}{4}\right) = \arctan\left(\frac{3}{4}\right) ]
Using a calculator to find ( \theta_2 ): [ \theta_2 \approx \arctan(0.75) \approx 36.87^\circ ]
The angles that the segments of the rope make with the horizontal are: