A wrench with a length of $15 c m$ $15 cm$ is used to unscrew a $\frac{5}{2} c m$ $5/2 cm$ bolt. If a torque of $15 N m$ $15 Nm$ is needed to overcome the friction keeping the bolt in place, what is the minimum torque that must be applied to the wrench to unscrew the bolt?

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A wrench with a length of $15 c m$ $15 cm$ is used to unscrew a $\frac{5}{2} c m$ $5/2 cm$ bolt. If a torque of $15 N m$ $15 Nm$ is needed to overcome the friction keeping the bolt in place, what is the minimum torque that must be applied to the wrench to unscrew the bolt?

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Answer 1 · 2017-08-19T23:43:25

$F = 100 N$ $F=100"N"$

Explanation:

I assume the question is meant to ask what is the minimum force required to unscrew the bolt.

Torque is given by:

$\Rightarrow τ = r F sin ϕ$ $=>color(blue)(tau=rFsinphi)$

Where $F$ $F$ is the applied force, $r$ $r$ is the distance from the pivot (axis of rotation) that the force is applied, and $ϕ$ $phi$ is the angle at which the force is applied (measured counterclockwise).

We can rearrange the equation to find the necessary force $F$ $F$ .

$\Rightarrow F = \frac{τ}{r sin ϕ}$ $=>color(blue)(F=tau/(rsinphi))$

We are given the following information:

$\mapsto r = 15 cm = 0.15 m$ $|->"r"=15"cm"=0.15"m"$
$\mapsto τ = 15 Nm$ $|->tau=15"Nm"$

It is implied that $ϕ = 90^{o}$ $phi=90^o$ since no angle—nor the necessary information to calculate one—is provided.

Note that I am not including the radius of the bolt in the measurement of the total radius because it should not affect the distance from the applied force to the bolt itself.

Therefore, we have:

$\Rightarrow F = \frac{15 Nm}{0.15 m}$ $=>F=(15"Nm")/(0.15"m")$

$\Rightarrow = 100 N$ $=>=100"N"$

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A wrench with a length of $15 c m$ $15 cm$ is used to unscrew a $\frac{5}{2} c m$ $5/2 cm$ bolt. If a torque of $15 N m$ $15 Nm$ is needed to overcome the friction keeping the bolt in place, what is the minimum torque that must be applied to the wrench to unscrew the bolt?

1 Answer

Explanation:

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1 Answer

Explanation:

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A wrench with a length of $15 c m$ $15 cm$ is used to unscrew a $\frac{5}{2} c m$ $5/2 cm$ bolt. If a torque of $15 N m$ $15 Nm$ is needed to overcome the friction keeping the bolt in place, what is the minimum torque that must be applied to the wrench to unscrew the bolt?