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sladkih [1.3K]
3 years ago
11

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.
Mathematics
1 answer:
Gemiola [76]3 years ago
8 0

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 .  =(0.1 \hat j) .(0\hat i+2 \hat j -4\hat k) =0+(0.1).2+0.(-4)=0.2

|\vec r|.|| 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.

                               

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