Question

A wrench 0.1 meters long lies along the positive yy-axis, and grips a bolt at the origin. A force is applied in the direction of ⟨0,2,−4⟩⟨0,2,−4⟩ at the end of the wrench. Find the magnitude of the force in Newtons needed to supply 100 Newton-Meters of torque to the bolt.

Answer 1

Answer:

Therefore the magnitude of the applied force is $500\sqrt 5$ N.

Step-by-step explanation:

Torque: Torque is the cross product of force and the distance of applied force  from the rotational axis.

$\tau=\vec F\times \vec r= |F||r| sin\theta$

Where $\theta$ is angle between F and r.

Newton-meter is the S.I unit of torque.

Along the positive y axis, the wrench lies .

The length of wrench is 0.1 m.

$\vec r= 0.1 \hat j$   [ since the vector $\vec r$ lies on y axis]

The direction of applied force is along the vector <0,2,-4>.

Then the angle between $\vec r$ and <0,2,-4> = the angle between F and $\vec r$.

We know that,

$\vec a.\vec b=|\vec a||\vec b|cos \theta$

$\vec r . <0,2,-4> =(0.1 \hat j) .(0\hat i+2 \hat j -4\hat k)$ $=0+(0.1).2+0.(-4)=0.2$

$|\vec r|.|<0,2,-4>| cos \theta= \sqrt{0.1^2}.\sqrt{0^2+2^2+(-4)^2} \ cos \theta$

$\therefore\sqrt{(0.1)^2}.\sqrt{0^2+2^2+(-4)^2} \ cos \theta= 0.2$

$\Rightarrow (0.1).2\sqrt5 cos \theta =0.2$

$\Rightarrow cos \theta =\frac{0.2}{0.2\sqrt 5}$

$\Rightarrow cos \theta =\frac{1}{\sqrt 5}$

We know that,

$sin \theta =\sqrt{1-cos^2\theta}=\sqrt{1-(\frac{1}{\sqrt5})^2}$

                            $=\sqrt {1-\frac15}$

                            $=\sqrt {\frac{5-1}{5}$

                            $=\frac{2}{\sqrt 5}$

Here $\tau$ = 100 N-m, $\vec r=0.1 \hat j$ and $sin \theta = \frac{2}{\sqrt 5}$

Now

$\tau=|F||r| sin\theta$

$\Rightarrow 100= |F|(0.1) \frac{2}{\sqrt 5}$

$\Rightarrow |F|=\frac{100\times \sqrt5}{0.2}$

$\Rightarrow |F|=500\sqrt 5$

Therefore the magnitude of the force is $500\sqrt 5$ N.