Question

Question #dfa0b

Answer 1

The correct answers are d and f.
The moment or torque is `0.02"Nm"` clockwise.
The arm is `4.9"cm"`.

Explanation:

The torque (or moment of force, angular force) is given by the formula
`vecM=vecr xx vecF`
where `vecr` is a vector from the pivot point to the location of the force (in this case from A to B), `vecF` is that force (in this case `vecW`).

This means the value of torque is
`M=rWsin(theta)`
where `theta` an angle beetwen vectors and direction is determined by right hand rule.
`theta=180^@-105^@=75^@` Actually this doesn't matter, since `sin(75^@)=sin(105^@)`

Extend the line of `vecW` and draw a line through A perpendicular to it.

This is equivalent to placing brackets like this
`M=W(rsin(theta))`

The segment labeled "y cm" is called arm. Its length is `rsin(theta)=5.1"cm"*0.966=4.926"cm"~~4.9"cm"` (f)

`M=0.33"N" * 4.9"cm"=1.62"N" * "cm"=0.0162"Nm"~~0.02"Nm"` (b,d)

`color(red)("Remember about units")`

The torque is clockwise for reasons obvious to me (The force is pulling right side down, so it will move down) (d)

Alternative method

Break `vecW` down to parallel and perpendicular component.

Parallel component `vec(F_2)` doesn't produce torque.
`vec(F_1)` is perpendicular to `vecr`, so
`M=rF_1`

This is equivalent to placing brackets like this
`M=r(Wcos(theta))`

One more method

`M=rWsin(theta)`

`r=5.1"cm"`, `W=0.33"N"` and `theta=75^@`

If all values are known, we can just calculate `sin(theta)` and multiply `rWsin(theta)`.

Answer 2

I agree with Voyager1: d and f.

Explanation:

I have one more method for calculating torque to add to the discussion.

Referring to my Physics book (by Resnick Halliday), the formula for Torque uses a vector cross product. To explain the vector cross product for the general case, my book says:

`vec C = vec A xx vecB = |A|*|B|*sinphi`

where `phi` is the angle between `vec A " and " vecB` measured in the manner that yields a value between `0^@ " and "180^@`.

When it discussed torque, my book defines it as
`vectau = vecr xx vecF`

Plugging in the data

`vectau = vecr xx vecF = 0.051 m*0.33 N*sin105^@ ~= 0.02 Nm`

I hope this helps,
Steve