DR. ZORRO HELP PLEASE?
1) Suppose the coefficient of kinetic friction between m A and the plane is μk= 0.11, and that m A = m B = 2.7kg .
Figure available at http://session.masteringphysics.com/problemAsset/1...
A) As m B moves down, determine the magnitude of the acceleration of m A and m B, given theta = 32 degrees.
B) What smallest value of μk will keep the system from accelerating?
2) If a curve with a radius of 90m is properly banked for a car traveling 67km/hr , what must be the coefficient of static friction for a car not to skid when traveling at 95 km/hr ?
- Dr. ZorroLv 78 years agoFavorite Answer
A) ( m_A + m_B) a = m_B g - m_A g sin(θ) - μ m_A g cos(θ)
As in my previous answer, the left is total inertia, right is net force.
Again, because the masses are equal, they cancel out:
2 a = g ( 1 - sin(θ) - μ cos(θ) )
a = ½ g ( 1 - sin(θ) - μ cos(θ) )
B) keeping the system from accelerating ( moving with a= 0) happens when
μk = ( 1 - sin(θ) ) / cos(θ)
2) properly banked, so the normal force alone has sufficient horizontal component to provide the centripetal force. Let's denote the properly banked speed by capital V and β the angle of the bank:
m V^2 / R = Fn sin(β)
m V^2 /R = m g cos(β) sin(β)
1/2 sin(2β) = V^2 / (gR)
sin(2β) = 2V^2 / (gR)
So the first part of the problem provides the information to calculate the angle β of the bank.
β = 1/2. arcsin(2V^2 / (gR))
Now the second part. Again we want to obtain a centripetal force for the higher speed v, this time provided by normal force component and by friction!
m v^2 / R = m g cos(β) sin(β) + μ m g cos(β) cos(β)
μ cos^2(β) = v^2/(gR) - sin(β) cos(β)
μ = v^2/(gR cos^2(β) ) - tan(β)
Use the calculated β from the first part.