Coefficient of the kinetic friction
μ₁ =0.3
Coefficient of the static friction
μ ₂ =0.5
m₁= 2.5 kg, m₂=2.5 kg
α=30°
(A)
m₁a= T- m₁gsinα- F(fr)… (1)
0=N-m₁gcosα………….(2)
m₂a= m₂g- T …………….(3)
F(fr)=μ₁N=μ₁•m₁gcosα
sum of (1) and (3):
a(m₁+m₂)=T- m₁gsinα -F(fr)- T+ m₂g=
= m₂g- m₁gsinα - μ₁m₁gcosα .
a=g(m₂-m₁sinα - μ₁m₁cosα)/(m₁+m₂) =…
T= m₂(g-a)
(B)
0= T- m₁gsinα- F(fr)… (1)
0=N-m₁gcosα………….(2)
0= m₂g - T …………….(3)
F(fr)=μ₂N= μ₂•m₁gcosα
(1) +(3)
0= T- m₁gsinα- F(fr)+ m₂g- T =
= m₂g- m₁gsinα- μ₂•m₁gcosα
m₁=m₂ =>
μ₂•cosα =1-sinα
Solve for α
A 2.5 kg block sits on an inclined plane with a 30 degree inclination. A light cord attached to the block passes up over a light frictionless pulley at the top of the plane and is tied to a second 2.5 kg mass freely hanging vertically. The coefficients of static and kinetic friction between the block and the plane are 0.5 and 0.3. When released from rest find: the acceleration of the blocks, the tension in the string, Explain why the tension supporting the hanging block is not equal to its weight and find the time for the block on the inclined plane to travel 0.5m up the plane. And find the minimum angle of inclination at which the block on the plane will remain at rest.
1 answer
1 answer