Question
"A 3.5 kg monket is suspended motionless 35 meter above the ground by holding two vines (a vertical vine and a diagonal vine)"
We need to find the tension force in the diagonal vine as well as the angle of the diagonal vine with respect to the y axis.
I'm having trouble making a motion map of this. However, I created one with two tension vectors, one up (I labeled it t2)and one up and diagonal(labeled t1) , a push going left, and weight (mass*gravity) going down. Therefore, my Max=t1sin45-17=0 and my max=t1costhetea+t2cos90-34.3=0. I did not get the right answers. But, I found t1 using my max equation then I used what I found in that to solve my may equation. I've become very lost.
Thanks for your responses!
We need to find the tension force in the diagonal vine as well as the angle of the diagonal vine with respect to the y axis.
I'm having trouble making a motion map of this. However, I created one with two tension vectors, one up (I labeled it t2)and one up and diagonal(labeled t1) , a push going left, and weight (mass*gravity) going down. Therefore, my Max=t1sin45-17=0 and my max=t1costhetea+t2cos90-34.3=0. I did not get the right answers. But, I found t1 using my max equation then I used what I found in that to solve my may equation. I've become very lost.
Thanks for your responses!
Answers
Steve
I don't see how one vine can be vertical and the other at an angle. If there's any tension in the side vine, it will move the "vertical" vine to the side.