Question

Mechanics-statics question

Answer 1

In the first part of the question, I added all the forces P,2P and 3P
Sum of Forces=3Pi+Pj+2Pk and in order to find the total moment I multiplied each force by the perpendicular distance from the moment point A
Sum of moments=3aPi+bPj+c2Pk

Answer 2

For the second part is this right?
Sum of moment/Sum of forces=d

Answer 3

@shamim

Answer 4

@Astrophysics

Answer 5

@AlexandervonHumboldt2

Answer 6

@IrishBoy123

Answer 7

i think your answers already look wrong, but i could of course be wring too!!

for the force, you have <3P, -P, 2P>

for torque, should you be recording them as vectors as well -- using the right hand thumb rule for direction - so i think you should get <Pb, -2Pc, 3aP> about that vertex?? and then you can try reducing that to zero to get some relationship between a,b,c

is there a reference point/ some background to this btw? a link or summat?

hope this helps rtaher than hinders.....

Answer 8

I am just given the forces in the x, y and z direction and their perpendicular distance to the moment point. I know that multiplying the force magnitude by the perpendicular distance will result in getting the magnitude of the moment.

Answer 9

not the moment components in the x, y and z directions.

Answer 10

have you been told to express the torques/moments as vectors pointing in direction of their axes of rotation?

Answer 11

yes i have been taught to express moments as vectors

Answer 12

ok

so specifically do you disagree with the conclusions i reached or are we in agreement on first bullet point in question....:p

Answer 13

okay i do agree with you that my answers were incorrect and that we must figure out the direction. Could I use matrix to find the moment caused by each force.

Answer 14

then add up all the moments of each force.

Answer 15

not sure i 'd go to that complication

eg, take force 3P. from (0,a,0), it creates a torque 3Pa. if you use the right hand thumb rule, that is a torque vector pointing up in the z direction so that makes it \(3Pa \; \hat k\) or \(3Pa \; \hat z\)

Answer 16

i got -pbi-2pcj and -3pak

Answer 17

@IrishBoy123 but shouldnt it be -a

Answer 18

curl your fingers in the direction of the moment about z and i think you will find your thumb should point upward, ie +ve z direction

Answer 19

I am not used into applying right hand rule. I used matrix instead.

Answer 20

If the system of forces is reduced to a single force, we should first reduce the given system of forces to a force couple system consisting of the resultan and couple vector M about A and this is possible only if they are perpendicular

Answer 21

I think they are coplanar

Answer 22

sorry for asking but you know the dot product?

Answer 23

nope.

Answer 24

well, i think we have this [if we don't, doesnt matter]

F = <3P, -P, 2P>

T = <Pb, -2Pc, 3aP>

if \(\vec F \perp \vec T\), you can boil this down to a single force. i think you said that above too.

so this happens if the dot prod = 0 cuz \(\mathbf{a}.\mathbf{b} = |\mathbf{a}|\mathbf{b}|| \cos \theta\)

So
\(\vec F \bullet \vec T = 0 \implies P<3,-1,2>\bullet P<b, -2c, 3a> = 0 \\ \implies 3b + 2c + 6a = 0\)

that's how i would do it, maybe you are being shown something different...

Answer 25

Thank you. Your discussion was most illuminating.
Just out of curiosity I performed the cross product F x (x,y,z) = T and when x, y and z are eliminated your condition on a, b and c unsurprisingly came out to be the same. However I will not do that again as the method was rather long winded.