Hello,

I imagine you are asking this question because you are looking for homework help or something of the sort.

This is the standard image for the problem that you are looking for. Your problem states that there are two ropes suspending the crate that has a mass of 100kg.

Firstly, it is important to convert the mass of the box into a weight, since it is being affected by gravity. This can be done by multiplying the box by gravity (mass*gravity = weight)

100kg * 9.81m/s^2 = 981N

Secondly, your problem states that there are two angles, being made from the vertical rather than the horizontal, meaning that the angle on the left makes a 70º angle with the horizontal (90º - 20º) and the angle on the right makes a 50º angle (90º - 40º) with the horizontal.

Once you have this information, you should then set up your system of equations.

For this problem, I will be doing it using the statics approach, since the sum of the forces acting on the object sum to zero. The primary equations are:

We will be using the equations for two dimensions.

∑Fx = 0 (The sum of the forces acting in the x-direction equals 0)

∑Fy = 0 (The sum of the forces acting in the y-direction equals 0)

First, we acknowledge that the weight of the crate is only acting in the y-direction, and has no effect on the x-direction, therefore, the x-direction will only be composed of the two tensions.

∑Fx = T1*cos(50º) - T2*cos(70º) = 0

Note that T2 is subtracted because the force of the tension is acting to the left, meaning that the force is not in the positive direction.

Simplifying, we get

T1 = 0.53209*T2

Next, we will sum up the forces acting in the y-direction.

∑Fy = T1*sin(50º) + T2*sin(70º) - 100(9.81) = 0

This time, the crate plays a part in how the force is applied in the y-direction, and because it is affected by gravity, its weight acts in the negative direction, while both of the vertical components of the tension act in the positive direction.

Instead of simplifying first, this time we will substitute the value of T1 into the equation so that the entire equation we be only with respect to the variable T2.

∑Fy = T1*sin(50º) + T2*sin(70º) - 100(9.81) = 0

Becomes

∑Fy = (0.53209*T2)*sin(50º) + T2*sin(70º) - 100(9.81) = 0

Which simplifies to

∑Fy = 0.4076*T2 + 0.9397*T2 - 100(9.81) = 0

Or

∑Fy = 1.3473*T2 = 981N

Solving for T2, we get the value

T2 = 728.125N

Now that we have the value for T2, we can simply plug it into the easier conversion formula we found a moment ago.

T1 = 0.53209*T2 = 0.53209*728.125N = 387.427N

T1 = 387.427N

General practice is to keep values within 3 significant figures, meaning that your answer is

T1 = 387N acting to the right

T2 = 728N acting to the left

Hopefully this helps!

~ Zach