A body of mass \(50 \mathrm{~kg}\) is suspended by two light, inextensible cables of lengths \(15 \mathrm{~m}\) and \(20 \mathrm{~m}\) from rigid supports placed \(25 \mathrm{~m}\) apart on the same level. Find the tensions in the cables. (Note that by convention 'light' means 'of negligible mass'. Take \(g=10 \mathrm{~ms}^{-2}\). This and the following two problems are applications of vector addition.)

There are three forces acting on the object: 1. The force of gravity, which acts downward and has magnitude equal to the mass times the gravitational acceleration: \(Fg = 50 \cdot 10 = 500 N\). This force acts at an angle of 90 degrees from the horizontal. 2. The tension force in the 15-meter cable, which we will call \(T1\). This force acts at an angle \(\theta_1\) from the horizontal. 3. The tension force in the 20-meter cable, which we will call \(T2\). This force acts at an angle \(\theta_2\) from the horizontal.

To find the angles \(\theta_1\) and \(\theta_2\), we can use the geometry of the problem. Since the cables are connected to supports 25 meters apart and the 15-meter cable is shorter than the 20-meter cable, we have a triangle with side lengths 15, 20, and 25 meters. Using the cosine rule, we can find the angle between the 15-meter and 20-meter sides: $$ cos(\theta_1) = \frac{15^2 + 25^2 - 20^2}{2 \cdot 15 \cdot 25} = \frac{225 + 625 - 400}{750} = \frac{450}{750} $$ Thus, \(\theta_1 = cos^{-1}\frac{3}{5}\). Similarly, we can find the angle between the 20-meter and 25-meter sides: $$ cos(\theta_2) = \frac{20^2 + 25^2 - 15^2}{2 \cdot 20 \cdot 25} = \frac{400 + 625 - 225}{1000} = \frac{800}{1000} $$ Thus, \(\theta_2 = cos^{-1}\frac{4}{5}\).

According to Newton's second law, the sum of the forces in each direction (horizontal and vertical) acting on the mass should be equal to zero, as the mass is in equilibrium. This can be represented by the following equations: Horizontal forces: $$ T1\cos(\theta_1) - T2\cos(\theta_2) = 0 $$ Vertical forces: $$ T1\sin(\theta_1) + T2\sin(\theta_2) - Fg = 0 $$

We now have a system of two equations with two unknowns, \(T1\) and \(T2\). Solving this system, we can find the values of \(T1\) and \(T2\) that satisfy the given conditions. Substituting the values of the angles and the gravitational force, we get: $$ T1\cos(cos^{-1}\frac{3}{5}) - T2\cos(cos^{-1}\frac{4}{5}) = 0 $$ $$ T1\sin(cos^{-1}\frac{3}{5}) + T2\sin(cos^{-1}\frac{4}{5}) - 500 = 0 $$ Since \(\cos(cos^{-1}(x)) = x\) and \(\sin(cos^{-1}(x)) = \sqrt{1 - x^2}\), we can rewrite the equations as: $$ \frac{3}{5}T1 - \frac{4}{5}T2 = 0 $$ $$ \frac{4}{5}T1 + \frac{3}{5}T2 - 500 = 0 $$ Multiplying the first equation by 3 and adding it to the second equation, we get: $$ T1 = \frac{500}{3} $$ Plugging this back into the first equation, we find: $$ T2 = \frac{4}{3}\cdot\frac{500}{3} $$ Calculating the tensions, we get: $$ T1 = \frac{500}{3} = 166.67 N $$ $$ T2 = \frac{2000}{3} = 666.67 N $$ So the tensions in the cables are approximately \(166.67 N\) in the 15-meter cable and approximately \(666.67 N\) in the 20-meter cable.