Question #dfa0b

Question

Question #dfa0b

2 Answers

Impact of this question

1227 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-12-23T18:51:35

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

Explanation:

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

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

Extend the line of $- \to W$ $vecW$ and draw a line through A perpendicular to it.
copied from above
This is equivalent to placing brackets like this
$M = W (r sin (θ))$ $M=W(rsin(theta))$

The segment labeled "y cm" is called arm. Its length is $r sin (θ) = 5.1 cm \cdot 0.966 = 4.926 cm \approx 4.9 cm$ $rsin(theta)=5.1"cm"*0.966=4.926"cm"~~4.9"cm"$ (f)

$M = 0.33 N \cdot 4.9 cm = 1.62 N \cdot cm = 0.0162 Nm \approx 0.02 Nm$ $M=0.33"N" * 4.9"cm"=1.62"N" * "cm"=0.0162"Nm"~~0.02"Nm"$ (b,d)

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

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 $- \to W$ $vecW$ down to parallel and perpendicular component.
copied from above
Parallel component $- \to F_{2}$ $vec(F_2)$ doesn't produce torque.
$- \to F_{1}$ $vec(F_1)$ is perpendicular to $\to r$ $vecr$ , so
$M = r F_{1}$ $M=rF_1$

This is equivalent to placing brackets like this
$M = r (W cos (θ))$ $M=r(Wcos(theta))$

One more method

$M = r W sin (θ)$ $M=rWsin(theta)$

$r = 5.1 cm$ $r=5.1"cm"$ , $W = 0.33 N$ $W=0.33"N"$ and $θ = 75^{\circ}$ $theta=75^@$

If all values are known, we can just calculate $sin (θ)$ $sin(theta)$ and multiply $r W sin (θ)$ $rWsin(theta)$ .

Answer 2 · 2017-12-23T21:37:38

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:

$\to C = \to A \times \to B = | A | \cdot | B | \cdot sin ϕ$ $vec C = vec A xx vecB = |A|*|B|*sinphi$

where $ϕ$ $phi$ is the angle between $\to A and \to B$ $vec A " and " vecB$ measured in the manner that yields a value between $0^{\circ} and 180^{\circ}$ $0^@ " and "180^@$ .

When it discussed torque, my book defines it as
$\to τ = \to r \times \to F$ $vectau = vecr xx vecF$

Plugging in the data

$\to τ = \to r \times \to F = 0.051 m \cdot 0.33 N \cdot {sin 105}^{\circ} ≅ 0.02 N m$ $vectau = vecr xx vecF = 0.051 m*0.33 N*sin105^@ ~= 0.02 Nm$

I hope this helps,
Steve

Science

Math

Humanities

... and beyond

Question #dfa0b

2 Answers

Explanation:

Explanation:

Related questions

Impact of this question