Recent content by cdornz

that definitely helped, thank you so much! I went to use this rule and apply it to another problem I had. The question for this one was if F1=50N and F2=10N and d1=9cm, and the distance between F2 and a is 2 times the distance between F1 and a, what is the torque about a? I set up total...

this is a basic example. I have other problems where there are two forces both going in that downward direction. I just want to make sure I am completing these problems correctly so I have a guide for the rest of my work down the line.

I know how to complete an actual torque equation and have done so, my question is more of what sign do I address the force of the problem. One problem I have is: F1=50N, d1=9cm, d2 = 10cm and d3=8cm F1 is situated between d2 and d3. I used the equation t=f(la) so my equation was...

In question with the signs, would the x-component be negative because it is technically below the F1 and the y-component positive because it is technically above the F1?

If the x-component is below the negative x-axis i would think that that component would be negative. But using those equations I think I've determined the answer. F1x=F1cos60° F1y=F1sin30° I broke this down by drawing out the picture of the triangle somewhat given and by finding the f1x...

So I was looking through my notes, and I saw the equation Vx = Vcosθ and Vy = Vsinθ. I'm not sure though that this is where i can put these equations to use, since i don't know the value of V. In looking at the angles, F1 is 60° from the x-axis going counterclockwise. I'm lost in more of how...

Being asked for the sum of forces in the X and Y directions and the axis are set up 'normally' so like a plus sign ( + ). I attempted the first one and I think I did it correctly, but I'm not so sure about the second one.

If "A" were a joint of the body, what is the compressive/tensive force at the joint? I already figured out the torque about A to be 6.840 Nm, but I'm not quite sure how to go about figuring out compressive/tensive forces. Normally, there would be a force associated in the area or angles...

thank you! and just one more question - slightly related to torque equations in general. When the force is exerted in that direction, it is considered a moment and therefore equal to 0 correct? Because no actual force is exerted on that line? So I would then think that the torque about p would...

This is what I worked it out to be, just want to make sure I'm doing it right.

Homework Statement What is the torque about a? F1 = 50N (located between d2 and d3) d1 = 9cm d2 = 10cm d3 = 8 cm Homework Equations T=(F)(Lever Arm) The Attempt at a Solution I'm extremely rusty with this material, but I realize I can't quite use the torque equation since the...

no, there are no angles or measurements given, which is why i thought I might of been over-analyzing the entire problem.

Assume the general position of x and y coordinates (looks like a plus sign, + ) break the vector down (draw) into x and y components.So does this mean that it wants me to draw just those coordinates around the vector it gives me? I have multiple problems like this and no sure if I am...

So I went with this and worked with the axes a bit to make it make more sense to me. I ended up with this as a result: \SigmaFx=max Tx+(mg)x = max T(cos100°) +mg(cos270°) = ma(cos180°) T(cos100°) = -ma T(cos100°) = -mv2/R T(sin100°) - mg = 0 T(sin100°)/g = m T(cos100°) =...

For components, I would have: Ftension; the y-component pointing upwards canceled out with the weight of the mass pointing downwards. There is no Ffriction since the object is in the air. Otherwise, I'm not sure what other forces. I could be going about this completely wrong, but from...