M1 Question

I'm really struggling to understand question 3a on the Solomon Paper L. I've looked at the markscheme but still can't understand how to get to the answer.
Could someone please explain the question and the markscheme?

They are just resolving the forces both vertically and horizontally.




First resolve horizontally, this will give, $T_Bcos(90- \alpha) = T_Acos\alpha$

Remember, $cos(90- \alpha) =sin\alpha$

So the equation becomes $\displaystyle T_Bsin\alpha = T_Acos\alpha \implies T_B \times \frac35 = T_A \times \frac45 \implies T_A=\frac{3T_B}{4}$

Now resolve vertically,
$T_Asin\alpha + T_Bsin(90-\alpha) = 1000g$

Remember, $sin(90- \alpha) =cos\alpha$

So the equation becomes, $\displaystyle T_Asin\alpha + T_Bcos\alpha = 1000g$

Now i think you can do it further.

Do you get it?

Ahhh, I didn't think C2 methods would be required for a M1 paper as the only prerequisite is C1, but it is a Solomon Press paper. Anyway, thanks for the help, I understand it now.

I actually would have done this question using a vector triangle, rather than the mark scheme technique.

Could you explain how to do it using a vector triangle?

See the image below, it is the vector triangle for it,




Now to find $T_B$, we will do,
$\displaystyle sin(90-\alpha) = \frac{T_B}{1000g} \implies cos\alpha = \frac{T_B}{1000g} \implies \frac45 \times 1000g = T_B \\ \implies \boxed{T_B=7840N}$

Do i make any sense?