Friction and Banked road question

[msq126]

Hello Everybody. I am having a very difficult time with these two questions.

The first question:

The system shown in Figure 4.54 has an acceleration of magnitude 1.77 m/s2, where m1 = 3.90 kg and m2 = 9.40 kg. Assume that the coefficient of kinetic friction between block and incline is the same for both inclines.


Figure 4.54
http://www.webassign.net/serpop/p4-54alt.gif

a)Find the coefficient of kinetic friction.?

b)Find the tension in the string. (N)

The Second Question:

A car rounds a banked curve as in the figure below. The radius of curvature of the road is R, the banking angle is θ, and the coefficient of static friction is μs.

http://www.webassign.net/serpop/5-13.gif

a)Determine the range of speeds the car can have without slipping up or down the banked surface. (Use theta for θ, mu for μs, R and g as necessary.)

vmin =

vmax=



b)Find the minimum value for μs such that the minimum speed is zero. (Use theta for θ, and R and g as necessary.)

μs =


c)What is the range of speeds possible if R = 100 m, θ = 10.0°, and μs = 0.110 (slippery conditions)?

vmin =


vmax=

-----------------------------

I would really appreciate it if someone could help me.
Thanks in Advance...

 

[tiny-tim]

Welcome to PF!

Hi msq126! Welcome to PF!

Hint: try good ol' Newton's second law for both questions.

Anyway, show us what you've tried, and where you're stuck, and then we'll know how to help.

 

[thomate1]

Hello, it seems like you are having some trouble with these questions. Let me try to provide some guidance and help you understand the concepts behind them.

For the first question, we need to use the equation for Newton's Second Law (F = ma) to find the coefficient of kinetic friction. We know that the system has an acceleration of 1.77 m/s2, and we have the masses of both blocks. We also know that the only force acting on the system is the force of friction, so we can set up the equation as follows:

F = μk * m1 * g = m1 * a

Where μk is the coefficient of kinetic friction, m1 is the mass of the first block, and g is the acceleration due to gravity (9.8 m/s2). Solving for μk, we get:

μk = (m1 * a) / (m1 * g) = a / g

Plugging in the values, we get μk = 1.77 / 9.8 = 0.1806. So, the coefficient of kinetic friction is 0.1806.

For part b, we need to find the tension in the string. We can use the same equation as before, but this time we will use the mass of the second block and the acceleration of the system.

F = μk * m2 * g = m2 * a

Solving for the tension, we get:

T = m2 * (a - μk * g) = 9.40 * (1.77 - 0.1806 * 9.8) = 13.91 N

So, the tension in the string is 13.91 N.

Moving on to the second question, we need to use the concept of centripetal force to find the range of speeds the car can have without slipping up or down the banked surface. The centripetal force is given by the equation Fc = mv2 / R, where m is the mass of the car, v is its speed, and R is the radius of curvature of the road.

At the minimum speed (vmin), the car is just about to slip up the banked surface. This means that the centripetal force is equal to the maximum static friction force (Ff = μs * mg), which prevents slipping. So, we can set up the equation as follows:

Fc