In our first example we will see that we can use a lever arm to greatly increase

Question

In our first example we will see that we can use a lever arm to greatly increase the torque without having to provide more force. Suppose, for instance, that an amateur plumber, unable to loosen a pipe fitting, slips a piece of scrap pipe (sometimes called a "cheater") over the handle of his wrench. He then applies his full weight of 900 N to the end of the cheater by standing on it. The distance from the center of the fitting to the point where the weight acts is 0.80 m, and the wrench handle and cheater make an angle of 19 with the horizontal (Figure 1) . Find the magnitude and direction of the torque of his weight about the center of the pipe fitting.

SOLUTION

SET UP (Figure 2) diagrams the position of the force exerted by the plumber in relation to the axis of rotation at point O.

SOLVE We have a choice of methods for solving this problem. From (Figure 2) , the angle between the vectors r  and F  is 109, and the moment arm l is

l=(0.80m)(sin109)=(0.80m)(sin71)=0.76m

=Fl=(900N)(0.76m)=680Nm

Or, from =rFsin,

=rFsin=(0.80m)(900N)(sin109)=680Nm

ALTERNATIVE SOLUTION Alternatively, we can start with the components of F . From (Figure 2) , we see that the component of force F  perpendicular to r  (which we call Ftan) is

Ftan=F(cos19)=(900N)(cos19)=850N

Then the torque is

=Ftanr=(850N)(0.80m)=680Nm

REFLECT The force tends to produce a counterclockwise rotation about O, so with the convention described above, this torque is +680 Nm, not 680 Nm. Also, please don't try to use this plumbing technique at home; you're very likely to break a pipe.

Part A - Practice Problem:

If the pipe fitting under consideration can withstand a maximum torque of 695 Nm without breaking, what is the smallest angle at which our amateur plumber can safely stand on the end of the cheater and apply his full weight without breaking the pipe fitting?

Express your answer in degrees to two significant figures.

In our first example we will see that we can use a lever arm to greatly increase the torque without having to provide more force. Suppose, for instance, that an amateur plumber, unable to loosen a pipe fitting, slips a piece of scrap pipe (sometimes called a "cheater") over the handle of his wrench. He then applies his full weight of 900 N to the end of the cheater by standing on it. The distance from the center of the fitting to the point where the weight acts is 0.80 m, and the wrench handle and cheater make an angle of 19 with the horizontal (Figure 1) . Find the magnitude and direction of the torque of his weight about the center of the pipe fitting.

SOLUTION

SET UP (Figure 2) diagrams the position of the force exerted by the plumber in relation to the axis of rotation at point O.

SOLVE We have a choice of methods for solving this problem. From (Figure 2) , the angle between the vectors r  and F  is 109, and the moment arm l is

l=(0.80m)(sin109)=(0.80m)(sin71)=0.76m

We can find the magnitude of the torque from either =Fl or =rFsin. From =Fl,

=Fl=(900N)(0.76m)=680Nm

Or, from =rFsin,

=rFsin=(0.80m)(900N)(sin109)=680Nm

ALTERNATIVE SOLUTION Alternatively, we can start with the components of F . From (Figure 2) , we see that the component of force F  perpendicular to r  (which we call Ftan) is

Ftan=F(cos19)=(900N)(cos19)=850N

Then the torque is

=Ftanr=(850N)(0.80m)=680Nm

REFLECT The force tends to produce a counterclockwise rotation about O, so with the convention described above, this torque is +680 Nm, not 680 Nm. Also, please don't try to use this plumbing technique at home; you're very likely to break a pipe.

Part A - Practice Problem:

If the pipe fitting under consideration can withstand a maximum torque of 695 Nm without breaking, what is the smallest angle at which our amateur plumber can safely stand on the end of the cheater and apply his full weight without breaking the pipe fitting?

Express your answer in degrees to two significant figures.

Explanation / Answer

angle made by the force with the handle = 90 - 19 deg

hence perpendicular component of the force F = sin(90 - 19) = cos(19)

hence

torque = F*r*cos(19)

F = 900 N

r = 0.8 m

hence

T = Frcos(19) = 680.77 Nm

A. Tmax = 695 Nm

let the smallest angle be phi

then

900*0.8*cos(phi) = 695

phi = 15.142 deg

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