Okay, a little tricky to understand, so any efforts will be appreciated!
A particle of mass 5Kg is on a rough horizontal surface. The coefficient between this particle and the surface is 'Mu'. This particle is connected to another, hanging vertically over a smooth peg of mass 3Kg. The acceleration of the system is 0.49m/s/s
Find the value of 'Mu'.
I'm definitely on the right lines, but i keep getting answers only close to the correct one. Please explain in depth how you got to your answer. Thanks in advance, Douglas
A particle of mass 5Kg is on a rough horizontal surface. The coefficient between this particle and the surface is 'Mu'. This particle is connected to another, hanging vertically over a smooth peg of mass 3Kg. The acceleration of the system is 0.49m/s/s
Find the value of 'Mu'.
I'm definitely on the right lines, but i keep getting answers only close to the correct one. Please explain in depth how you got to your answer. Thanks in advance, Douglas
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Hello
the net force on the system is the weight force of the 3 kg mass minus the friction force of the
5kg mass.
Fnet = m1g - m2gμ = 3*9.81 - 5*9.81*μ
the acceleration is Fnet/(m1+m2)
0.49 = (3*9.81 - 5*9.81*μ)/8
3.92 = 29.43 - 49.05μ
μ = 0.52 <-- ans.
Regards
the net force on the system is the weight force of the 3 kg mass minus the friction force of the
5kg mass.
Fnet = m1g - m2gμ = 3*9.81 - 5*9.81*μ
the acceleration is Fnet/(m1+m2)
0.49 = (3*9.81 - 5*9.81*μ)/8
3.92 = 29.43 - 49.05μ
μ = 0.52 <-- ans.
Regards
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IIt is not stated where the peg is, so I will assume ( a pure guess) that it is a horizontal distance from the first mass so the the second mass hangs below the edge of the surface.
The force of gravity on the second mass is an unbalanced force and it accelerates the system. The acceleration is opposed by the inertia of the masses and the friction of the second mass.
The inertia is accounted for with F = (m1 + m2) * a, where F will be the weight of mass2 (m2*g)
The frictional force is determined by f = mu * N , where N is the weight of mass1 (m1*g)
F tends to accelerate the mass, f tends to retard the acceleration. These forces are in opposite directions so should be subtracted.
F(net) = F - f
(m1 + m2)*a = m2*g - mu*m1*g
(5 + 3)*0.49 = 3*9.8 - mu*5*9.8
3.92 = 29.4 - 49*mu
mu = (29.4 - 3.92) / 49
mu = 0.52
The force of gravity on the second mass is an unbalanced force and it accelerates the system. The acceleration is opposed by the inertia of the masses and the friction of the second mass.
The inertia is accounted for with F = (m1 + m2) * a, where F will be the weight of mass2 (m2*g)
The frictional force is determined by f = mu * N , where N is the weight of mass1 (m1*g)
F tends to accelerate the mass, f tends to retard the acceleration. These forces are in opposite directions so should be subtracted.
F(net) = F - f
(m1 + m2)*a = m2*g - mu*m1*g
(5 + 3)*0.49 = 3*9.8 - mu*5*9.8
3.92 = 29.4 - 49*mu
mu = (29.4 - 3.92) / 49
mu = 0.52