Question 2. A car is driving around a circular loop of radius R. The coefficient of friction between the tires and the road is µ. Initially, the car is travelling with speed vo, but at t = 0 the driver begins to accelerate. If the magnitude of the acceleration is a, how long will it take for the car to begin to slip and thus deviate from the circular path? How far will the car have traveled by that point?
Question 2. A car is driving around a circular loop of radius R. The coefficient of friction between the tires and the road is µ. Initially, the car is travelling with speed vo, but at t = 0 the driver begins to accelerate. If the magnitude of the acceleration is a, how long will it take for the car to begin to slip and thus deviate from the circular path? How far will the car have traveled by that point?
Related questions
Question
Please answer question 2, answer 1 is in the 2nd image
Transcribed Image Text:Question 1.
A mover is pulling a box with a force of magnitude F directed at an angle of a above the horizontal. Assuming
the friction coefficient is u, what value of a maximizes the acceleration of the box? Hint: sin(x) = cos(x)
and cos(x) = – sin(x)
Question 2.
A car is driving around a circular loop of radius R. The coefficient of friction between the tires and the
road is u. Initially, the car is travelling with speed vo, but at t = 0 the driver begins to accelerate. If the
magnitude of the acceleration is a, how long will it take for the car to begin to slip and thus deviate from
the circular path? How far will the car have traveled by that point?
Transcribed Image Text:7:09 PM Tue Feb 23
* 84%
88
00
Dynamics_4 -
Arch History: Global Architectures Of Belief And Authority
Dynamics_4
Question 1.
A mover is pulling a box with a force of magnitude F directed at an angle of a above the horizontal. Assuming
the friction coefficient is µ, what value of a maximizes the acceleration of the box? Hint: sin(x) = cos(x)
and cos(r) = - sin(x)
M
Mg
O M: mass
O Taking the force components alang the vertical,
The box is at equilibrium along the vertical direction
EF,
N- Mg
+ Fsin d=0
N: Mg - Fsnd li)
2 O Here N is the
normas contact force,
Then the
Kinetic friction
can be
.calculated.as ,
fx =LN
: M (Mg - fond).
is Fret = Fcos d-fix
- FLos h - H Ma -Fsina)
Along the
hori zontal
direction, the
net force
3
O using the second law
of motion
Fnet : Ma
a= fnet
M
- Fros a -H(Mg - Fsin d),
M
H (Mg -Fsind)
Frosd
M
M
Frosd
uF sind
M
M
4) For Max accelecation da = O
da = 0
Fras d _ M Mg +H Fsin d
M
M
M
H F sind'
d Fosh
dcd l M
M )
da
M
Ě d (cosa)
M dh
-0. +HF d.(sind)=0
M da
El-sina) +hE Cosa)-0
M
- Sind + M CUS cd=0
Sin d > H COSd
fandh = M
d= tan
Therefore to maxini ze the acceleration the reavired
angle is d= tan" (M)
Expert Solution
Step 1
(2)
If the centripetal force is obtained only by the force of friction (Ff) between the tires of the vehicle and road, then for safe turn the maximum velocity be given as,
That means after this velocity the car will start slipping.
Trending now
This is a popular solution!
Step by step
Solved in 4 steps
