Recent content by SirMarksAlot

Wouldnt the force anywhere still be 48 N since the couple cancels each other out, and leaves only the moment? Also, is the line of action just the direction/line that the force follows?

Homework Statement Also, I am not entirely sure what the question is exactly asking. Also, what does the line of action represent? Homework Equations M = Fx(Perpendicular distance) + Fy(Perpendicular distance) For couples: M=Fd The Attempt at a Solution i was thinking...

oh wow, all these question are so easy now, thanks for the advice about the reciprocal.

kk got it. RS =(1 + secθ) / (secθ - 1) =(1 + 1/cosθ) / (1/cosθ - 1) =([cosθ+1] / cosθ) / ([1-cosθ] / cosθ) =[(cosθ+1)(cosθ)] / [(1-cosθ)(cosθ)] cancel cosθ =(1 + cosθ) / (1 - cosθ)

Homework Statement (1 + cosθ) / (1 - cosθ) = (1 + secθ) / (secθ - 1) Homework Equations using only the quotient identities, pythagorean identities, and reciprocal identities The Attempt at a Solution didnt know where to start...

I don't know how to find the time it would take to reach the max height.

oh wouldn't the final velocity be 0 when it reaches a maximum height? because he would have stopped moving up?

when u say x = vcosθ*t , what is v? the speed before the jump?

honestly, I am struggling and have no idea what to do...

do u mean these formulas? Fx = cos A * F1 Fy = sin A * F1

but all i have is the angle of the jump and the distance...no time or anything

honestly, i have no idea

how would i go about solving the velocity for each component?

kk thank you, no wonder i kept getting the wrong answer

Homework Statement A runner jumps at 30 degrees to the ground and covers 8.90 m. What was the takeoff speed? The Attempt at a Solution I have no idea where to start...