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For reference, this is Question 18 from page 64 of Mechanics and Probability -

https://books.google.co.uk/books?id=NjeaS1rcFrIC&pg=PA64&lpg=PA64&dq=the+resultant+of+2P+and+Q+is+also+a+force+of+magnitude+P&source=bl&ots=jPX_9GuKIU&sig=YGFK6dKprC7axfMK-ZQJesCIvM8&hl=en&sa=X&redir_esc=y#v=onepage&q=the%20resultant%20of%202P%20and%20Q%20is%20also%20a%20force%20of%20magnitude%20P&f=false -

Forces P and Q act along lines OA and OB respectively and their resultant is a force of magnitude P. If the force P along OA is replaced by a force 2P along OA, the resultant of 2P and Q is also a force of magnitude P. Find:

a) The magnitude of Q in terms of P

b) The angle between OA and OB

c) The angles which the two resultants make with OA

I tried the following but I can't see where to go:

Let A and B be arbitrary points in the first quadrant of the Cartesian plane. Let the x and y components of the forces P and Q be $P_x,P_y,Q_x$ and $Q_y$ respectively, then:

$\sqrt{(P_x+Q_x)^2+(P_y+Q_y)^2} = \sqrt{(2P_x+Q_x)^2+(2P_y+Q_y)^2}$

I was hoping maybe to find an expression for $\sqrt{Q_x^2+Q_y^2}$

but it just leads to:

$3(P_x^2+P_y^2)=-2P_yQ_y-2P_xQ_x$

Any help greatly appreciated, Thanks, Mitch.

--------- EDIT ---------

Thanks Ofek Gillon for your answer to parts A and B. I've solved part C as below but I feel like I'm missing a simpler route:

Here's my solution:

Let the angle between the first resultant P+Q and OA be $\theta_1$ and the angle between the second resultant 2P+Q and OA be $\theta_2$

Then we have:

Resultant Angles To OA

$(P+Q).P = |P+Q||P|cos\theta_1$

$\therefore$ $P^2+Q.P=\frac{1}{\sqrt3}|Q|Pcos\theta_1$

call this equation (2)

Similarly

$(2P+Q).P=|2P+Q||P|cos\theta_2$

$\therefore 2P^2+Q.P=\frac{1}{\sqrt3}|Q|Pcos\theta_2$

call this equation (1)

$\therefore$ (1) - (2) and substituting for Q gives:

$P^2=|P|P(cos\theta_2-cos\theta_1)$

$\therefore cos\theta_2-cos\theta_1=1$

call this equation (5)

Resultant Angles To OB

$(P+Q).Q=|P+Q||Q|cos(\frac{5\pi}{6}-\theta_1)$

$\therefore P.Q+Q^2=\sqrt3|P|Pcos(\frac{5\pi}{6}-\theta_1)$

Call ths equation (4)

Similarly

$(2P+Q).Q=|2P+Q||Q|cos(\frac{5\pi}{6}-\theta_2)$

$\therefore 2(P.Q)+Q^2=\sqrt3|P|Pcos(\frac{5\pi}{6}-\theta_2)$

Call this equation (3)

$\therefore$ (3) - (4) gives:

$|P||Q|cos\frac{5*\pi}{6}=\sqrt3|P|P[cos(\frac{5\pi}{6}-\theta_2)-cos(\frac{5\pi}{6}-\theta_1)]$

$\therefore \frac{\sqrt3}{2}=cos(\frac{5\pi}{6}-\theta_1)-cos(\frac{5\pi}{6}-\theta_2)$

Call this equation (6)

So we need to solve (5) and (6) simultaneously.

Well in equation (6) if we set $cos(\frac{5\pi}{6}-\theta_1)=\frac{\sqrt3}{2}$ we get $\theta_1=\frac{2\pi}{3}$

But of course we must also have $cos(\frac{5\pi}{6}-\theta_2)=0$ which gives $\theta_2=\frac{\pi}{3}$

We see that these two values will simultaneously satisfy equation (5) and so we finally have verified that:

$\theta_1=120^{\circ}$ and $\theta_2=60^{\circ}$

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  • $\begingroup$ You assume that $A$ and $B$ can both be chosen in the first quadrant, but it isn't clear that you can do that. In particular it would limit the angle between $OA$ and $OB$ in a way that you've not shown is necessarily justified. $\endgroup$ – hardmath Mar 16 '17 at 1:07
  • $\begingroup$ Thanks, you are quiet right. $\endgroup$ – gnitsuk Mar 16 '17 at 15:24
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A)

Forces P and Q act along lines OA and OB respectively and their resultant is a force of magnitude P.

$$(I) \ \ \ \ \ (P+Q)\cdot (P+Q) = P^2 $$ $$(I) \ \ \ \ \ 2P\cdot Q + Q^2 =0 $$

the resultant of 2P and Q is also a force of magnitude P.

$$(II) \ \ \ \ \ (2P+Q)\cdot (2P+Q) = P^2 $$ $$(II) \ \ \ \ \ 4P^2 + 4P\cdot Q + Q^2 = P^2 $$

Subtracting $(II) - 2(I)$ we get $$4P^2 - Q^2 = P^2 $$ $$Q^2 = 3P^2$$ $$ |Q| = \sqrt{3} |P|$$


B) Substituting that in $(II)$:

$$ 4P^2 + 4P\cdot Q + 3P^2 = P^2$$ $$ 6P^2 = -4P\cdot Q$$ $$ 3|P|\cdot |P| = -2 |P| |Q| \cos \theta $$ $$ 3|P| = -2\cdot \sqrt{3} |P| \cos \theta $$ $$ \cos \theta = -\frac{\sqrt{3}}{2} $$ $$\theta = \frac{5\pi}{6}$$


C) After I introduced the algebraic and geometric definitions of the dot product, try mixing the two in this section. Good luck!

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  • $\begingroup$ Thank you for your answer. At this stage in the book the concept of the dot product has not been introduced. I wonder if the question is mistakenly present or whether an anwser in terms of more primitive constructs is intended. Thanks again, Mitch. $\endgroup$ – gnitsuk Mar 16 '17 at 9:21

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