Recent content by newbphysic

Thank you for your help :bow:

P is fixed but Q vary. Ooh i see so is this mean in order to find minimum value the unknown must perpendicular to each other ?

I'm really confused :bugeye: Why for the first problem i can't solve it like this one (without substituting FA) FA sin(20) -FB sin(q) = 0 FB =FAsin(20)/sin(q) I know that we don't know FA but we know in order for FB to be minimum sin(q) must equal 1 (the biggest value) but the result will not...

when it's perpendicular to FR ? the red line is the right answer according to calculation :oldconfused: OK :wink: but what i still don't understand is this, how can you know it must be perpendicular to FA not to FR. Here is another example that looks exactly like before but in this example Fb...

For FB perpendicular FA,is there an explanation for this ? why it works ? and why not perpendicular to FR ? Sorry if i ask a lot of questions :)

950 sin(q) + 2626.47 cos(q) = 0 950 tan (q) = -2626.47 tan q = - 2626.47 / 950 q =

70.11 clockwise

so it's just a coincidence that it has the same result ?

if i use that method the result is the same. q = 70 degrees clockwise from x.

All right, thanks a lot ehild. :smile::smile::smile: One more question though, there is other method that says in order for Fb to be minimum it must perpendicular to Fa ? Can you explain why it works ?

950 sin(q) + 2626.47 cos(q) = 0

my guess q must be 135(-225) or 315(-45) because sin and cos have different sign there

numerator must be zero which means q must be 0 for sine and 90 for cosine

The answer : :nb) hmm,it looks like mission impossible . Do you know another method ehild ?

derivative of cos and sin will result another sin and cos what should i do with it?