If FB = 2kN and the resultant force acts along

If $F_B = 2kN$ and the resultant force acts along the positive u axis, determine the magnitude of the resultant force and the angle u.

Solution:

Let us first draw the vector components as follows:

Notice that the resultant is along the u-axis as stated by the question.

Now, we will draw the vector components tail to tail as follows:

We can now use law of sines to figure out $\phi$ . (Forgot the law of sines?)

$\frac{\sin \phi}{3}=\frac{\sin30^0}{2}$

$\phi=48.69^0$

From this, we can figure out $\theta$ .

$\theta=30^0+\phi$

$\theta = 78.59^0$

Now we can use the law of sines again to figure out $F_R$ .

$\frac{F_R}{\sin(180^0-78.59^0)}=\frac{2}{\sin30^0}$

$F_R=3.92kN$

This question can be found in Engineering Mechanics: Statics (SI edition), 13th edition, chapter 2, question 2-7.

2 thoughts on “If FB = 2kN and the resultant force acts along”

Reply ↓

m-1999 July 29, 2018 at 2:47 am

Is there any another way to slove it ? and if there is how we can do it ? Thanks

Reply ↓

questionsolutions Post author July 29, 2018 at 7:00 am

I am sure there are other ways to solve these types of questions though I haven’t utilized them. Most textbooks usually show two ways of solving these questions, either the tip to tip method or the parallelogram way. Please take a look: https://goo.gl/ssjJMP

Many thanks and best of luck with your studies. 🙂

If FB = 2kN and the resultant force acts along 2

Solution:

Let us first draw the vector components as follows:

Notice that the resultant is along the u-axis as stated by the question.

Now, we will draw the vector components tail to tail as follows:

We can now use law of sines to figure out $\phi$ . (Forgot the law of sines?)

$\frac{\sin \phi}{3}=\frac{\sin30^0}{2}$

$\phi=48.69^0$

From this, we can figure out $\theta$ .

$\theta=30^0+\phi$

$\theta = 78.59^0$

Now we can use the law of sines again to figure out $F_R$ .

$\frac{F_R}{\sin(180^0-78.59^0)}=\frac{2}{\sin30^0}$

$F_R=3.92kN$

This question can be found in Engineering Mechanics: Statics (SI edition), 13th edition, chapter 2, question 2-7.

Leave a comment Cancel reply

2 thoughts on “If FB = 2kN and the resultant force acts along”

Solution:

Let us first draw the vector components as follows:

Notice that the resultant is along the u-axis as stated by the question.

Now, we will draw the vector components tail to tail as follows:

We can now use law of sines to figure out \phi. (Forgot the law of sines?)

\frac{\sin \phi}{3}=\frac{\sin30^0}{2}

\phi=48.69^0

From this, we can figure out \theta.

\theta=30^0+\phi

\theta = 78.59^0

Now we can use the law of sines again to figure out F_R.

\frac{F_R}{\sin(180^0-78.59^0)}=\frac{2}{\sin30^0}

F_R=3.92kN

This question can be found in Engineering Mechanics: Statics (SI edition), 13th edition, chapter 2, question 2-7.

Leave a comment Cancel reply

2 thoughts on “If FB = 2kN and the resultant force acts along”

We can now use law of sines to figure out $\phi$ . (Forgot the law of sines?)

$\frac{\sin \phi}{3}=\frac{\sin30^0}{2}$

$\phi=48.69^0$

From this, we can figure out $\theta$ .

$\theta=30^0+\phi$

$\theta = 78.59^0$

Now we can use the law of sines again to figure out $F_R$ .

$\frac{F_R}{\sin(180^0-78.59^0)}=\frac{2}{\sin30^0}$

$F_R=3.92kN$