Yes, that is the correct equation. F = mg + mv^2/R

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The discussion focuses on calculating the forces acting on a biology student riding a bike around a corner. The relevant equation for total force is confirmed as F = mg + mv^2/R. Given the student's mass of 87 kg, speed of 8.3 m/s, and corner radius of 24 m, the calculations yield a total force of approximately 1108.5 N. The coefficient of static friction is also discussed, with a minimum value of 0.292 established to prevent skidding. The conversation emphasizes the importance of understanding the forces involved in circular motion.
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A biology student rides her bike around a corner of radius 24 meter at a steady speed of 8.3 m/sec. The combined mass of the student and the bike is 87 kg. The coefficent of static friction between the bike and the road is μs = 0.39.

a) If she is not skidding, what is the magnitude of the force of friction on her bike from the road?
Ffric = N *
--------------------------------------------------------------------------------
b) What is the minimum value the coefficient of static friction can have before the bike tire will skid?
μmin = *
0.292 OK
--------------------------------------------------------------------------------
c) What is the magnitude of the total force between the bike tire and the road?
Ftotal = N
1103.19 NO

Need help with part c.

Is it something like this F = mg + mv^2/R

Is this correct.
 
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Naeem said:
A biology student rides her bike around a corner of radius 24 meter at a steady speed of 8.3 m/sec. The combined mass of the student and the bike is 87 kg. The coefficent of static friction between the bike and the road is μs = 0.39.

a) If she is not skidding, what is the magnitude of the force of friction on her bike from the road?
Ffric = N *
--------------------------------------------------------------------------------
b) What is the minimum value the coefficient of static friction can have before the bike tire will skid?
μmin = *
0.292 OK
--------------------------------------------------------------------------------
c) What is the magnitude of the total force between the bike tire and the road?
Ftotal = N
1103.19 NO

Need help with part c.

Is it something like this F = mg + mv^2/R
The total force on the bicycle is:

\vec F = m\vec g + \vec N + \mu_sN\hat r = m\vec a = \frac{mv^2}{R}\hat r

The forces exerted by the road on the bicycle are \vec N, \mu_sN\hat r. The magnitude of those forces would be the magnitude of the vector sum:

m\vec g - \frac{mv^2}{R}\hat r = \vec N + \mu_sN\hat r

Note\vec N and \hat r are perpendicular. What is the magnitude of the resultant force?

AM
 


Yes, that is the correct equation for calculating the total force between the bike tire and the road. To solve for it, we can plug in the given values into the equation:

F = mg + mv^2/R
= (87 kg)(9.8 m/s^2) + (87 kg)(8.3 m/s)^2/24 m
= 852.6 N + 255.9 N
= 1108.5 N

So the magnitude of the total force between the bike tire and the road is approximately 1108.5 N.
 
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