A wrench with a length of #42 cm# is being used to unscrew a #7/2 cm# bolt. If a torque of #32 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
Aug 10, 2017

Answer:

#F~~76"N"#

Explanation:

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

Torque is given by the equation:

#color(darkblue)(tau=rFsinphi)#

where #F# is the applied force, #r# is the distance from the pivot (axis of rotation) that the force is applied, and #phi# is the angle at which the force is applied.

We can rearrange the equation for torque to find #F#.

#color(crimson)(F=tau/(rsinphi))#

We are provided the following information:

  • #|->"r"=42"cm"=0.42"m"#

  • #|->tau=32"Nm"#

And it is implied that #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 I am assuming that this is the radius of the body of the bolt and not the head of the bolt.

Therefore, we have:

#F=(32"Nm")/(0.42"m")#

#=76.19"N"#

#~~76"N"#