# Thread: Slider crank

1. ## Slider crank

Is anyone able to point me in the right direction to answering these questions please.

2. ## Re: Slider crank

Surely you can solve b). Point a) can be solved with a simple application of the Pythagoras theorem.

In general, let P' be the projection of P on OR. If you know angle POR, then you an find OP' and PP' using trigonometric functions. Knowing PP', you can find P'Q using the Pythagoras theorem. Therefore, you can find OQ.

3. ## Re: Slider crank

Thank you. I will try to review trignomometric functions.

4. ## Re: Slider crank

Is the answer for (b) = 3.25 - 1.5 = 1.75

5. ## Re: Slider crank

Originally Posted by arangu1508
Is the answer for (b) = 3.25 - 1.5 = 1.75
1. What you have calculated is the distance $|\overline{OQ}|$ if P is placed on the left end of the diameter of the circle.

2. As far as I understand the question you are asked to determine the length of the way the point Q is traveling on the line OR if the point P describes a semicircle. And that must be a complete diameter, that means the answer is 1.5 m.

6. ## Re: Slider crank

Thank you once again earboth, with your help I was able to understand the question. Hopefully I have answered them correctly (although you have already confirmed b)

a) Initially Q is @ 3.2 m after P rotates 90 degress using the pythagoras theorem we find

$c^2=b^2-a^2$
$sqrt(2.5^2-0.75^2)= 2.38 m$

Distance the end Q travelled = 3.2 m - 2.38 m = 0.82 m

b) Distance the end Q travelled (half a turn) 2 x .75 = 1.5 m

c) using the sine rule we find sin(30) c = 3.12 m
using the sine rule again we find sin(170) c =1.76 m

Distance the end Q travelled 1.36 m

If I have made any mistakes please let me know. Thank you.

7. ## Re: Slider crank

Originally Posted by Dcoz
a) Initially Q is @ 3.2 m after P rotates 90 degress
It's 3.25 m, so the traveled distance in a) is 0.87 m, not 0.82 m.

Originally Posted by Dcoz
c) using the sine rule we find sin(30) c = 3.12 m
using the sine rule again we find sin(170) c =1.76 m

Distance the end Q travelled 1.36 m
I don't understand this. First, what is c? 3.12 / sin(30) is not the same as 1.76 / sin(170). Second, sin(30) determines the vertical position of P, not the horizontal position of Q.

As I said, one way is to find PP' and OP' (where P' is the projection of P) and then use Pythagoras theorem to find P'Q; then OQ = OP' + P'Q.

8. ## Re: Slider crank

Originally Posted by Dcoz
Thank you once again earboth, with your help I was able to understand the question. Hopefully I have answered them correctly (although you have already confirmed b)

a) Initially Q is @ 3.2 m after P rotates 90 degress using the pythagoras theorem we find

$c^m2=b^2-a^2$
$sqrt(2.5^2-0.75^2)= 2.38 m$

Distance the end Q travelled = 3.2 m - 2.38 m = 0.82 m

b) Distance the end Q travelled (half a turn) 2 x .75 = 1.5 m

c) using the sine rule we find sin(30) c = 3.12 m
using the sine rule again we find sin(170) c =1.76 m

Distance the end Q travelled 1.36 m

If I have made any mistakes please let me know. Thank you.
1. I'll take emakarov's suggestions. I only use some different labels. (see attachment. I've modified your image a little bit)

2.
• $|\overline{OF}| = p = 0.75 \cdot \cos(\theta)$
• $|\overline{PF}| = 0.75 \cdot \sin(\theta)$
• $|\overline{FQ}| = q = \sqrt{2.5^2-(0.75 \cdot \sin(\theta))^2}$

The traveled distance w:

$w = 3.25 - (p+q)$

3. The advantage of emakarov's method is that the traveled distance w depends only on the angle of rotation $\theta$. So you only have to plug in the angle to get the distance from the start-point.

9. ## Re: Slider crank

Thank you both for your help on this one. I'm not 100% certain I have got the answer to c) right

this is what I have

ϴ = 30˚ p + q = 0.6495 + 2.848 = 3.1336
ϴ = 170˚ p + q = -0.7386 + 2.4954 = 1.756
3.1336 – 1.756 = 1.38 m

Thank you.