Asked by Kampamba Nsofu
A 5kg object placed on a friction-less, horizontal table connected to a string that passes over a pulley and then is fastened to a hanging 9kg object.
(A) Draw a free body diagram of both the objects.
(B) The Coefficient of kinetic friction between the object and the surface is 0.35 and the coefficient of static friction is 0.46. Find: (i)The acceleration of the two objects
(ii)The tension in the cable
(A) Draw a free body diagram of both the objects.
(B) The Coefficient of kinetic friction between the object and the surface is 0.35 and the coefficient of static friction is 0.46. Find: (i)The acceleration of the two objects
(ii)The tension in the cable
Answers
Answered by
Anonymous
T = spring tension
WHEN NO FRICTION:
Object 1
force up from table = 5 g but irrelevant, balanced by gravity 5 g down
horizontal force = T
so
T = 5 a
Object 2
force down = 9 g
force up = T
so
9 g - T = 9 a
WITH Friction
Object 1
Horizontal force =
T - mu (5 g) = 5 a
Object 2
9 g - T = 9 a
======================
so add two equations to eliminate T
9 g - 5 mu g = 14 a
a = (g/14)(9-5 mu)
If mu < 9/5 it moves so use mu = .35 , the sliding coeffficient
a = (9.81/14)(9 - 5*.35)
now use a to go back and get
T=9(9.81-a)
WHEN NO FRICTION:
Object 1
force up from table = 5 g but irrelevant, balanced by gravity 5 g down
horizontal force = T
so
T = 5 a
Object 2
force down = 9 g
force up = T
so
9 g - T = 9 a
WITH Friction
Object 1
Horizontal force =
T - mu (5 g) = 5 a
Object 2
9 g - T = 9 a
======================
so add two equations to eliminate T
9 g - 5 mu g = 14 a
a = (g/14)(9-5 mu)
If mu < 9/5 it moves so use mu = .35 , the sliding coeffficient
a = (9.81/14)(9 - 5*.35)
now use a to go back and get
T=9(9.81-a)
Answered by
Girma
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