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

I CORRECTIn summary, we have a biology student riding her bike around a corner with a radius of 24 meters at a constant speed of 8.3 m/sec. The combined mass of the student and bike is 87 kg and the coefficient of static friction between the bike and the road is 0.39. If the bike is not skidding, the magnitude of the force of friction on the bike from the road is unknown. The minimum value of coefficient of static friction before the bike tire will skid is 0.292. In order to find the magnitude of the total force between the bike tire and the road, we use the equation F = mg + mv^2/R. The total force exerted by the
  • #1
Naeem
194
0
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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  • #2
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:

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

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

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

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

AM
 
  • #3


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.
 

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

1. What does the equation F = mg + mv^2/R represent?

The equation F = mg + mv^2/R represents the total force acting on an object. The first term, mg, represents the force due to gravity, while the second term, mv^2/R, represents the force due to centripetal acceleration.

2. Why is it important to use this equation?

This equation is important because it allows us to calculate the total force acting on an object, taking into account both gravity and centripetal acceleration. This is crucial in many scientific and engineering applications, such as calculating the forces on a satellite in orbit or the forces on a roller coaster.

3. What do the variables in this equation stand for?

The variable F represents force, m represents mass, g represents the acceleration due to gravity, v represents the velocity of the object, and R represents the radius of the circular path the object is traveling on.

4. Is this equation always applicable?

No, this equation is only applicable in situations where there is both gravity and centripetal acceleration acting on the object. For example, if an object is only under the influence of gravity and not moving in a circular path, this equation would not be applicable.

5. How is this equation derived?

This equation can be derived using Newton's Second Law of Motion, which states that force is equal to mass times acceleration. In this case, the acceleration is the sum of the acceleration due to gravity and the centripetal acceleration.

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