Force acts on a pebble with position vector , relative to the origin. What is the resulting torque acting on the pebble about (a) the origin and (b) a point with coordinates (2.61 m, 0, -8.03 m)?

Incomplete question, but let analyse generally a torque acting on a rotating body

Explanation:

Toque is given as

τ = r × F

Where

τ is the torque (Nm)

r is the position vector (m)

F is the force (N).

Let assume we have the following data

F= x•i + y•j + z•k

Also let assume the position vector is

r'= a•i + b•j + c•k

a. The first question, torque at the origin

The origin will have a position vector of r1(0,0,0)

Then, the displacement is

r=AB=AO+OB=-OA+OB

r = r'-r1 = (a, b, c)-(0,0,0)

r=(a, b, c)

Then, the torque is given as

τ = r×F

τ = (a, b, c) × (x, y, z)

Note that

i×i=j×j=k×k=0

i×j=k, j×i=-k

j×k=i, k×j=-i

k×i=j, i×k=-j

τ = (a•i + b•j + c•k) × (x•i + y•j + z•k)

τ = a•i×(x•i + y•j + z•k) + b•j×(x•i + y•j + z•k) + c•k×(x•i + y•j + z•k)

τ = (a•i × x•i)+ (a•i × y•j) + (a•i × z•k) + (b•j × x•i) + (b•j × y•j) + (b•j × z•k) + (c•k × x•i) + (c•k × y•j) + (c•k × z•k)

Now, using the above law.

τ = 0 + ay•k - az•j - bx•k + 0 + bz•i + cx•j - cy•i + 0

τ= ay•k - az•j - bx•k + bz•i + cx•j -cy•i

Then, rearranging

τ = (bz - cy)•i + (cx - az)•j + (ay - bx)•k

This is general formula for the torque at the origin, so you can apply it to the given values in your question because the question given is not complete.

b. Now let, analyse at the position vector (2.61 m, 0, -8.03 m)

Using the same analysis

position vector of r1(2.61, 0, -8.03)m

Then, the displacement is

r=AB=AO+OB=-OA+OB

r = r'-r1 = (a, b, c)-(2.61,0,-8.03)

r=(a-2.61, b, c+8.03)

Then, the torque is given as

τ = r×F

τ = (a-2.61, b, c+8.03) × (x, y, z)

Note that

i×i=j×j=k×k=0

i×j=k, j×i=-k

j×k=i, k×j=-i

k×i=j, i×k=-j

τ =([a-2.61]•i, b•j, [c+8.03]•k) × (x•i + y•j + z•k)

τ = (a-2.61)•i×(x•i + y•j + z•k) + b•j×(x•i + y•j + z•k) + (c+8.03)•k×(x•i + y•j + z•k)

τ = ((a-2.61)•i × x•i)+ ((a-2.61)•i × y•j) + ((a-2.61)•i × z•k) + (b•j × x•i) + (b•j × y•j) + (b•j × z•k) + ((c+8.03)•k × x•i) + ((c+8.03)•k × y•j) + ((c+8.03)•k × z•k)

Now, using the above law.

τ = 0 + (a-2.61)y•k - (a-2.61)z•j - bx•k + 0 + bz•i + (c+8.03)x•j - (c+8.03)y•i + 0

τ= (a-2.61)y•k - (a-2.61)z•j - bx•k + bz•i + (c+8.03)x•j -(c+8.03)y•i

Then, rearranging

τ = (bz - cy - 8.03y)•i + (cx + 8.03x - az - 2.61z)•j + (ay - 2.61y - bx)•k

This is general formula for the torque at the point (2.61 m, 0, -8.03 m), so you can apply it to the given values in your question because the question given is not complete.

the force on the thing and the distance the force moves the thing

one person pulls 90n one way and the other has 100n pulling the other way.

so the net force = 90 - 100 = -10n,

or the ans can be 10n if the picture shows -90n and 100n instead.