You're driving your pickup truck around a curve that has a radius of 17 m.
1 answer:
Answer:
Explanation:
The centripetal force experienced by the steel toolbox is provided by the friction between the toolbox and the bed of the truck.
Therefore,
--------------------------------(1)
Here,
v - speed of truck
m - mass of toolbox
r - radius of curve around which the truck is turning
f - frictional force
μmg
In order to obtain Vmax, frictional force must also be max
Thus,
f = μmg--------------------------------------------------(2)
Substitute (2) in (1)
μmg
μgr
g - 9.8 m/
The standard value of coefficient of static friction between steel and steel is
μ ≈ 0.7
Therefore,
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and this weight is applied at the center of mass of the meterstick, so at x=0.50 m, therefore at a distance
from the pivot.
The torque generated by the weight of the meterstick around the pivot is:
To keep the system in equilibrium, the mass of 0.50 kg must generate an equal torque with opposite direction of rotation, so it must be located at a distance d2 somewhere between x=0 and x=0.40 m. The magnitude of the torque should be the same, 0.20 Nm, and so we have:
from which we find the value of d2:
So, the mass should be put at x=-0.04 m from the pivot, therefore at the x=36 cm mark.
and this weight is applied at the center of mass of the meterstick, so at x=0.50 m, therefore at a distance
from the pivot.
The torque generated by the weight of the meterstick around the pivot is:
To keep the system in equilibrium, the mass of 0.50 kg must generate an equal torque with opposite direction of rotation, so it must be located at a distance d2 somewhere between x=0 and x=0.40 m. The magnitude of the torque should be the same, 0.20 Nm, and so we have:
from which we find the value of d2:
So, the mass should be put at x=-0.04 m from the pivot, therefore at the x=36 cm mark.