The maximum torque a bolt can withstand without breaking is 100 N m. A mechanic tightens the bolt with a spanner 0.25 m long. What is the ..
maximum force the mechanic can use on the end of the spanner?
If he used a shorter spanner, would he be more likely or less likely to break the bolt? Give reasons for your answer.
Torque of a force about an axis is equal to magnitude of the force force multiplied by distance of line of application of the force from the axis.
It is given that the force is applied at a distance of 0.25 meters. Therefor the torque T to which the bolt is subjected to equal to:
T = 0.25 x f
Where f = force applied
It is given that the maximum permissible is torque T(max) is equal to 100 N.
Therefore T(max) = 100 = 0.25 x f(max)
Where f(max) is the maximum permissible force.
Therefore: f(max) = 100/0.25 = 400 N
Maximum force the mechanic can use at the end of spanner is 400 N.
In case the mechanic uses a spanner shorter than 0.25 m. a greater force may be applied. The exact maximum force that can be applied is given by the formula:
f(max) = 100/(Length of spanner in m)
The force causes a translatory motion in a body and similarly torque causes the rotatory motion. In physics it is also called the moment of the force about a point or pivot. The magnitude of torque is the product of the force and the perpendicular distance of the pivot from the force. So , if F is the force applied at angle x to the arm of length r of the spanner, about a pivot (here the bolt), then the torque T = r * Fsinx and is expressed in units od Newton meter. So, the torque the mechanic can exert through the spanner arm length r=0.25 m, which the bolt could sustain is given by:
100Nm > 0.25 F sinx, or F < 100Nm/(0.25m sinx) = 400 N/sin x
So at 90 degree, sinx = 1.Threfore, F should be within 400 N.
If uses his force, at angle 70 degree to the spanner, he can use a focce still higher but less than 400/sin70 = 425.67 N.
If he uses the spanner of lesser arm length say, 0.20m, then
it requires higher force to produce the required torque and the force is given by: F < 100Nm/(0.20m*sin x) = 500N/sinx. So it is less likely that he applies higher force which exhuasts him. Still a lower arm of spanner ,say 0.10m , if he utilises, it requires, a force F < 100Nm/(0.10m*sinx) =1000N/sinx , which is still more difficult and strenuous for the mechanic to exert and he is still less likely to break the bolt.