what is the magnitude of the torque about the point ?

Question

a partical is located at the vector position 2.2m + 4m and the force acting on it is 3.4N + 2.7N. what is the magnitude of the torque about the point having coordinates [2m,9m] ?

Answer 1

First, calculate the distance from 2.2,4 to 2,9
=sqrt(.2^2+5^2)
=5 m
This is the moment arm.
The torque of the forces(I will assume that the particle has no mass) is the force that is perpendicular to the line intersecting the two points times the moment arm (distance)

The line from the point to the particle has the following equation
y=-25x+59

A line perpendicular to this line subtends an angle of
atan(59/2.3)
=87.7 degrees w/r/t the x axis

This could be solved by using vector crossing, I'm a bit rusty with the matrix approach, so I will look at each component individually

I will set counterclockwise torque as positive

The vertical is 2.7 N up

cos(87.7)*2.7*5
T=-.526

The horizontal is 3.4 to the right
=sin(87.7)*3.4*5

T=17

these sum to
=17.5

Sorry for the multiple posts, I had the particle at the wrong point. It helps to sketch a picture.

j

Answer 2

FIRST I propose (if U have not accomplished this) sketching a diagram of the beam and the forces that act on it If U try this there is the question of the place does the load of the beam act? the respond IF THE BEAM IS OF UNIFORM bypass-area is on the midpoint (a million.4 m from the tip of the beam). with the intention to me this very needed actuality is lacking from this question's description of the beam. Assuming beam to be uniform. torque approximately attachment end of weight of beam = a million.4 x 330(9.80 one) = 4532 m-N torque approximately attachment end of weight of worker = 2.8 x seventy two.0(9.80 one) = 1978 m-N entire torque approximately attachment end of all weights = 6510 m-N upload'l: I write contraptions of torque/2d interior the SI gadget as meter-Newtons {m-N}. {this is backwards, yet does not actual rely, from the Newton-meters {N-m} frequently written while fixing artwork/means probs.