Asked by mark
Let the radius of the circular segment be
108 m, the mass of the car 1861 kg, and
the coefficient of the static friction between
the road and the tire 0.9. The banking angle
is not given.
Find the magnitude of the normal force N which the road exerts on the car at the opti-mal speed (the speed at which the frictional
force is zero) of 72 km/h.
Answer in units of N.
108 m, the mass of the car 1861 kg, and
the coefficient of the static friction between
the road and the tire 0.9. The banking angle
is not given.
Find the magnitude of the normal force N which the road exerts on the car at the opti-mal speed (the speed at which the frictional
force is zero) of 72 km/h.
Answer in units of N.
Answers
Answered by
Sorab
FBD: friction down angled road/ramp, normal force perpendicular to ramp, gravity down, impending motion up ramp.
Fnet = ma
In r direction,
Friction_r+N_r=mv^2/r
(f_s)(cosX)+NsinX=mv^2/r
uNcosX+NsinX=mv^2/r
N=(mv^2)/(ucosX+sinX) (1)
In z direction, NcosX-uNsinX-W=0
N=mg/(cosX-usinX) (2)
Set equations 1 and 2 equal, solve for X, substitute back into either 1 or 2, yielding N
Fnet = ma
In r direction,
Friction_r+N_r=mv^2/r
(f_s)(cosX)+NsinX=mv^2/r
uNcosX+NsinX=mv^2/r
N=(mv^2)/(ucosX+sinX) (1)
In z direction, NcosX-uNsinX-W=0
N=mg/(cosX-usinX) (2)
Set equations 1 and 2 equal, solve for X, substitute back into either 1 or 2, yielding N
Answered by
Sorab
lol, Equation 1 is actually
N=(mv^2)/(r)(ucosX+sinX)
N=(mv^2)/(r)(ucosX+sinX)
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