A mass m1 = 10 kg on top of a rough horizontal table surface is connected by a massless cable over a frictionless wheel to a hanging mass m2 = 5 kg, as shown in the previous problem. In m2 falls 1 m from rest in 1.2 seconds, find the co-efficient of kinetic friction between m1 and the table surface.
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A mass m1 = 10kg resting on a rough horizontal table surface is connected by a massless cable over a frictionless wheel to a hanging mass m2= 5 kg,. What is the minimum co-efficent of static friction which will allow the masses to remain at rest?
1.
h=at²/2
a=2h/t²= 2•1/1.2²=1.4 m/s²
m2•a =m2•g –T = > T = m2•g-m2•a.
m1•a=T-F(fr) = T-μ•N= T-μ•m1•g = m2•g-m2•a- μ•m1•g.
μ=[m2•g – a(m1+m2)]/m1•g = [5•9.8 – 1.4•15]/10•9.8 = 0.27.
2.
0 = m2•g –T
0=T-F1(fr) = T- μ(s) •m1•g
m2•g = μ(s) •m1•g,
μ(s) = m2/m1=5/10 = 0.5
To find the coefficient of kinetic friction between m1 and the table surface, we can use the equations of motion.
1. Find the acceleration of m2:
Since m2 falls 1 m from rest in 1.2 seconds, we can use the equation of motion:
s = ut + 0.5at^2
Substituting the given values:
1 = 0.5a(1.2)^2
1 = 0.5a(1.44)
a = 1.39 m/s^2
2. Find the tension in the cable:
Using Newton's second law, ∑F = ma, for m2:
T - m2g = m2a
Substituting the values:
T - (5 kg)(9.8 m/s^2) = (5 kg)(1.39 m/s^2)
T - 49 = 6.95
T = 55.95 N
3. Find the frictional force:
Since there is no vertical acceleration for m1, the vertical component of the tension balances the weight:
m1g = T sin θ
Substituting the values:
(10 kg)(9.8 m/s^2) = 55.95 N sin θ
4. Find the horizontal force due to friction:
The horizontal component of the tension provides the horizontal force:
F_friction = T cos θ
Substituting the values:
F_friction = 55.95 N cos θ
5. Find the normal force:
Since m1 is resting on a horizontal table, the normal force is equal to m1g:
Normal force = m1g = (10 kg)(9.8 m/s^2) = 98 N
6. Calculate the coefficient of kinetic friction:
Using the equation of friction: F_friction = μ_k * Normal force
Substituting the values:
55.95 N cos θ = μ_k * 98 N
Simplifying:
μ_k = (55.95 N cos θ) / 98 N
μ_k = 0.57 cos θ
Therefore, to find the coefficient of kinetic friction, we need to determine the angle θ.
To find the minimum coefficient of static friction that will allow the masses to remain at rest, we can use the concept of equilibrium.
1. Find the tension in the cable:
Using Newton's second law, ∑F = 0, for m2:
T - m2g = 0
Therefore, T = m2g = (5 kg)(9.8 m/s^2) = 49 N
2. Find the maximum frictional force:
The maximum frictional force is given by:
F_friction_max = μ_s * Normal force
Since we are looking for the minimum coefficient of static friction, we can assume that friction is at its maximum.
3. Find the normal force:
The normal force is equal to m1g:
Normal force = m1g = (10 kg)(9.8 m/s^2) = 98 N
4. Calculate the minimum coefficient of static friction:
Using the equation of friction:
F_friction_max = μ_s * Normal force
Substituting the values:
μ_s * 98 N = 49 N
Simplifying:
μ_s = 49 N / 98 N
μ_s = 0.5
Therefore, the minimum coefficient of static friction that will allow the masses to remain at rest is 0.5.