Math Help Forum: Slider crank

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

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.

Thank you. I will try to review trignomometric functions.

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

Originally Posted by arangu1508

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

1. What you have calculated is the distance 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.

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




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.

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.

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




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.

• 
  
• 
  
• 

The traveled distance w:



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

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.