Recent content by mmaismma

Yes you are right. There is an exception for such small boards that I couldn't figure out. What is a 'ceil'?

You are right but I didn't simplify the formula intentionally. Because this way one can easily figure out how I have derived the formulas.

In my answer I have already told that required number is always maximum in case 2 except for ##m=1## which will give ##9## for the given case. However if you want me to distinguish the cases as ##m=1,m=2,m>2##, here it is.

If my answer is correct part (b) is very simple. -----

I have solved part (a). I will post the answer in text form when I solve both parts. Untill then I have numbered all the equations. -----

Summary:: The set of values of ##b## for which the origin and the point ##(1, 1)## lie on the same side of the straight line ##a^2x+aby+1=0## ##\forall~a\in\mathbb{R},~b>0##.(a) ##a\geq1## or ##a\leq-3## (b) ##a\in~(-3,~0)\cup(\frac13,~1)## (c) ##a\in~(0,~1)## (d) ##a\in~(-\infty,~0)## I tried...

Yeah you are right. \begin{align}&=mg\frac{1+3\cos^2\theta}4\nonumber\\&=mg\frac{1+3-3\sin^2\theta}4\nonumber\\&=mg\frac{4-3\sin^2\theta}4\nonumber\\&=\mathbf{mg(1-\frac{3sin^2\theta}4)}\end{align}

You are right. Continuing previous equation: ##\begin{align}&=mg\frac{1+3\cos^2\theta}4\nonumber\\&=mg\frac{1+3-3\sin^2\theta}4\nonumber\\&=mg\frac{4-4\sin^2\theta}4\nonumber\\&=\mathbf{mg(1-\frac{sin^2\theta}4)}\end{align}## And...

You are right my bad. But it still doesn't match my answer key: ##\begin{align}&N_x\cos\theta+N_y\sin\theta=N_1\nonumber\\&mg\cos\theta\times\cos\theta+\frac{mg\sin\theta}4\sin\theta\nonumber\\&\frac{mg(4\cos{}^2\theta+\sin{}^2\theta)}4\neq N_1\nonumber\end{align}##

You are right it ain't striking me. I can't understand what are you trying to say.

What did't strike me? Since I have got to the answer, You can now post the complete answer as per Physics forums rules.

$$\begin{align}N_1&=N_y\cos\theta+N_x\cos\theta\nonumber\\ &=\frac{3mg\sin\theta\cos\theta}4+mg\cos{}^2\theta\neq\frac{3mg\sin\theta\cos\theta}4\nonumber\end{align}$$

It doesn't match my anwer key. Probably the answer key is wrong. But can you ensure me if my answer key is wrong. Here is what my answer key says: $$N_1=\frac{3mg\sin\theta\cos\theta}{4}\\ N_2=mg\left[1-\frac{3\sin{}^2\theta}{4}\right]$$

Sorry. My bad. Here is my revised attempt. Please check. ## \begin{align} \Rightarrow F_{net}{}_{x} &=ma_x\nonumber\\ \Rightarrow mg\sin \theta - N_y &= ma_t \nonumber\\ \Rightarrow N_y &= mg\sin \theta - ma_t \nonumber\\ &= mg\sin\theta -m \times \frac {3g\sin\theta} {2l} \times \frac {l} {2}...