A 40kg mass hangs from two cables. One cable is at a 25 degree angle. The other cable is horizontally attatched to the mass. Find the tension for each cable

Fm = M*g=40kg * 9.8N/kg = 392 N. = Force of the mass.

T1*Cos25 = -T2.
T2 = -T1*Cos25 = -0.91T1.

T1*sin25 = 392.
T1 = 928 N.

T2 = -0.91*T1 = -0.91 * 928 = -841 N.

To find the tension for each cable, we can use the principles of equilibrium. In equilibrium, the sum of all the forces acting on an object is zero.

Let's define the following variables:
T1 = tension in the cable at the 25 degree angle
T2 = tension in the horizontally attached cable
θ = angle between the horizontal and the horizontally attached cable

Now, let's analyze the forces acting on the 40kg mass:

1. Weight force (mg):
The weight force acts vertically downward and its magnitude is given by:
mg = 40kg × 9.8m/s^2 = 392 N

2. Tension T1:
This tension force is acting at an angle of 25 degrees. We can break this tension force into vertical and horizontal components:
T1*sin(25°) = T2, since the vertical component of T1 balances the weight force.
T1*cos(25°) = T1

3. Tension T2:
This tension force is acting horizontally. It can only balance the horizontal component of T1.

Now, let's write the equations based on the equilibrium condition:

In the vertical direction:
T1*sin(25°) = mg

In the horizontal direction:
T1*cos(25°) = T2

We can solve these equations simultaneously to find the values of T1 and T2.

First, let's solve the vertical equation:
T1*sin(25°) = 392 N
T1 = 392 N / sin(25°)
T1 ≈ 928.08 N

Now, let's substitute the value of T1 into the horizontal equation:
928.08 N * cos(25°) = T2
T2 ≈ 838.12 N

Therefore, the tension in the cable at the 25 degree angle (T1) is approximately 928.08 N, and the tension in the horizontally attached cable (T2) is approximately 838.12 N.