I really can not that you enough for this perfectly clear explanation! I finally understand it now, many thanks for writing it in such a clean and detailed paragraph, it allowed me to see exactly in which places i was getting stuck. I very much appreciate it. Have a great day and thank you again!
Oh my god i finally get it thank you so much, so the tension in the cable is 73.5? and then in order to find the tension in each one of the cables should i use trigonometry and do cos(30)=73.5/x, x being my tension in the cable linked to the leash?
if my tension force is equal doesn't that mean that together they make 0 and thus there is no force acting upon the sled?
Is there any way for me to find the forces of tension on the knot without finding the force of tension in the cable pulling the sled?
So my friction force is 73.5, if i use the equation f= mu x N, mu being 0.05 and N 1471.5N (9.81 x 150kg). So my sum of forces would be Ft - 73.5, Ft being my unknown tension force, and if i try to put that into Newton's second law equation it gives me
Ft - 73.5 = 150 x 0 so Ft = 73.5 but that...
Sorry about that, i meant the force i am looking for so the tension force in the cable. i was hoping to get that by finding the sum of forces with Newton's second law and then taking the friction force out of the sum
If i use my acceleration as 0 though my force will be 0 horizontally and i thought the objective was to find my force horizontally to find the tension force in the cable.
i thought my vertical acceleration is non zero because there is a force of gravitation puling me down and i assumed that...
So in this case it is as i doubted, i can not use the 2nd law because my acceleration is 0 in every direction except y (perpendicular to the ground), and the F that i find if the law is applied to that direction is not the one i am looking for since i need the F horizontally (x axis). That's why...
Right, i think what you're reffering to is the normal force? But the normal force compensates for the gravitational force so does that mean my sum of forces on the Z axis is equal to 0?
About this, if i am going at constant speed my acceleration is 0 because otherwise the speed would keep...
So on the z axis, the only force that exists is my gravitational force, so that would give us Fz=M x Az which is Fz = 150 x 9.81 = 1471.5N for my force on the Z axis right?
My frictional force points in the opposite ditection of the rope tension so that would mean that the only resulting force is my tension force except it is a bit diminished because of the friction right?
Sorry for my bad drawing but i represented it first from a side pov and then as viewed from above. The problem is i do not understand how to apply it to each component individually Because the acceleration of each is still 0 and that would lead to F=MA. If i divide in 3 with Fx, Fy and Fz then...