Math Problems dealing with Torque

General Concept Problem

A bolt on a car engine needs to be tightened with a torque of 35Nm. You use a 25cm long wrench and pull on the end of the wrench at an angle of 60 degrees from the perpendicular. How long is the lever arm? How much force do you have to exert?


Known:

r= 0.25m
Torque= 35Nm
Θ=60

Unknown :
L= ?
F= ?

Solve for L :
L=r sin Θ
L=(0.25m)(sin 60)
L= 0.22m

Solve for F:
Torque=Force(sin Θ)
Force= Torque/r sin Θ
Force= 35Nm/(0.25m)(sin60)
Force= 1.6 x 102

See-Saw Problem

Kariann (56 kg) and Aysha (43 kg) want to balance on a 1.75c long seesaw. Where should they place the pivot point?

Known:
Mk= 56kg
M A= 43 kg
Rk + RA = 1.75m

Unknown:
Rk= ?
RA=?

Find the two forces.

Kariann:
FgK=mKg
FgK=(56 kg) (9.8 m/s2)
FgK=5.5 x 102N

Aysha:
FgA=mAg
FgA=(43 kg) (9.8m/s2)
FgA=4.2 x 102

Define Kariann’s distance in terms of the length of the seesaw and Aysha’s distance.

RK= 1.75m- rA

Where there is no rotation the sum of the torques is equal to zero.

Fgkrk=FgArA
(Fgk)rk- (FgA)rA=0Nm

Fgk (1.75m - rA) - FgArA =0Nm

Solve for rA

Fgk (1.75m) - FgK (rA) - FgArA = 0Nm
(FgK) rA+ (FgA)rA= Fgk (1.75m)
rA (FgK+FgA)= Fgk (1.75m)
rA= Fgk (1.75m)/FgK+FgA
rA= (5.5 x 102N) (1.75m) / (5.5 x 102N + 4.2 x 102 N)
rA=0.99m

Ladder Problem

What is the maximum distance x from a smooth frictionless wall you can put a 100N uniform ladder of length 5.0m on a floor where the coefficient of static friction is 0.40?




Solution: http://zebu.uoregon.edu/~probs/mech/circ/Torquilibrium3/node2.html