huh?! *confused*
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this looks harder than what i normally do >.<
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first question is JC physics with friction added only
2nd question need to consider force
In short, you need to draw out all the forces to analyse.
Try first... Cannot do tmr I do... my last paper tmr... after wait for my first class cert only :D
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ok :s
add: wah cert x)
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okay.. i lost some marks for the first question.. still wrong :x
im not sure about the second question
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mass times friction coefficient to test slipping... or is it force. it should be mass :x ahh wth
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help D:?
the force pulling the block up is the same as the one going down?
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far out im tank.
derived
acceleration of the block = a = (mg + uk Mgcos # - Mgsin# )/(M+m) and the T = mg - ma
# = the given angle
M = the greater mass
m = the smaller mass
uk = the given kinetic coefficient of frictionso answer for first bit should be 281.6126N
okay second question...
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for q2:
Us = g sin T / (w^2 r + g cos T)
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Q1
Assume it is moving already... If it is not, we have to redo with μs... So now we do with μk
Let the tension in the string be T
For the 54kg block,
Reaction force is 54g cos 42
Downward force is 54g sin 42Taking moving downwards left as positive,
54 * a = 54g sin42 - T - μk 54g cos42
54a = 315.0979 - T -- (1)For the 27kg block,
upward force = T
downward force = 27g
Taking upwards as positive,
T - 27g = 27a
27a = T - 264.87
54a = 2T - 529.74 -- (2)(2) - (1): 3T - 844.8779 = 0
T = 281.626N -
Q2
Reaction force = mg cos 54 + mrω^2 = m * (5.766 + 18.28224) = 24.0484 m
let coefficient of static friction be μs
μs * 24.0484 m = mg sin 54
μs = 0.33 -
thanks, confirms what ive done is correct