0
$\begingroup$

I am trying to get my head around some introductory mechanics, however I need some help.

For this question:

Number 4

It asks me to resolve the $200kN$ force shown acting on the pin into components in the (a) $x$ and $y$ directions, (b) $x’$ and $y$ directions and (c) $x$ and $y’$ directions.

Now, I know for part (a) $R_{x}=200\cos{40}=153.21$ (2 d.p.) and $R_{y}=200\sin{40}=128.56$ (2 d.p.).

However, for parts (b) and (c), isn't it true that $R_{x'}=200\cos(40+30)$ and $R_{y'}=200\sin(40+30)$?

Also, I don't understand why $R_{x}$ and $R_{y}$ change when you are resolving them with respect to $R_{x'}$ and $R_{y'}$ and not each other.

$\endgroup$
0
$\begingroup$

No. The problem does not ask you to write $R$ in terms of $R_{x'}$ and $R_{y'}$. Forget everything you did in part a. What the problem asks in part b is to find two forces, one along $x'$ the other along $y$, such as the sum of them is $R$. You can write $R_{x'}$ as made up of a component along $x$: $R_{xx'}=R_{x'}\cos(30)$ and along $y$: $R_{yx'}=-R_{x'}\sin(30)$. Now write $R$ along $x$ and $y$ and you get $$R\cos(40)= R_{x'}\cos(30)$$ and $$R\sin(40)=R_y-R_{x'}\sin(30)$$ Equivalently, you can project the required components along $R$ and the direction perpendicular to $R$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.