# The towing pendant AB is subjected to the force 6

The towing pendant AB is subjected to the force of 50 kN exerted by a tugboat. Determine the force in each of the bridles, BC and BD, if the ship is moving forward with constant velocity. Image from: Hibbeler, R. C., S. C. Fan, Kai Beng. Yap, and Peter Schiavone. Statics: Mechanics for Engineers. Singapore: Pearson, 2013.

#### Solution:

We will start by drawing the free body diagram like so: ##### Sometimes, it is harder to understand a question when the orientation of the free body diagram is not to our typical style. In this question, you can turn the free body diagram to better suit your understanding like so: ##### The free body diagram still represents the same forces, and it makes it easier for us to visually understand the question.

We will choose forces going $\rightarrow^+$ to be positive and $\uparrow+$ to be positive.

Let us now write the equations of equilibrium for the x-axis forces.

$\sum \text{F}_\text{x}=0$

$\mathrm{T_{2}}\sin 30^0-\mathrm{T_{1}}\sin 20^0=0$

Now, we will write the equations of equilibrium for the y-axis forces.

$\sum \text{F}_\text{y}=0$

$\mathrm{T_{2}}\cos 30^0+\mathrm{T_{1}}\cos 20^0-50=0$

We can now solve both equations simultaneously to find $T_1$ and $T_2$ like so:

Isolate for $T_2$

$T_2=\dfrac{\mathrm{T_{1}}\sin 20^0}{\sin 30^0}$

Substitute $T_2$ into the second equation

$(\dfrac{\mathrm{T_{1}}\sin 20^0}{\sin 30^0})\cos 30^0+\mathrm{T_{1}}\cos 20^0-50=0$

$T_1=32.6$ kN

Now that we know what $T_1$ is, substitute this value back into our previous equation.

$T_2=\dfrac{{32.6}\sin 20^0}{\sin 30^0}$

$T_2=22.2$ kN

## 6 thoughts on “The towing pendant AB is subjected to the force”

• João Victor

Nice post, but i have a question, why in the equilibrum of the y-axis forces you wrote (T2*cos30º – T1*cos20º…)? Those forces are in the same direction when projected on the y-axis, they should be added, shouldn’t they?

• questionsolutions Post author

You are absolutely right! It was a typo on our end, though the final answers are correct. Thank you very much for letting us know of the mistake, it has been fixed 🙂

• SUBHANUDDIN

when the reference axis is y, then we will use sin instead of cos in the summation of x axis forces. and vice versa.
AM I RIGHT???

• questionsolutions Post author

Yes, in this case, it would flip around, however, you shouldn’t think of it that way. Instead, quickly imagine a right angle triangle and use the sin/cos to break a force into components. That way, even if opposite angles are given, it will still work out.

• Mateen

Hi, can you please tell me why did you choose for x axis sin and for y axis sin. Normally we choose x for cos theta and y for sin theta. it is confusing here.

• questionsolutions Post author

Hi, it’s a case by case situation. Please don’t think that it’s always cos for x and sin for y. This is simply not true, it depends on how the angle is given to us. You should imagine a right angle triangle to get the right answer. Let’s look at cable T2. We are given the angle of 30 degrees. Sine is opposite over hypotenuse. So opposite in this case is the x-component of T2. Cosine is adjacent over hypotenuse. Adjacent in this case is the y-component of T2. If you wanted to use x for cos and y for sin, you need to use the other angle, which is 90-30 = 60 degrees. In every question, imagine a right angle triangle and then break it into components. It will guarantee your answer is right, if you always assume x for cos and y for sin, you might get some questions wrong. 🙂