huh?! confused

WoAiMeiMei SG

this looks harder than what i normally do >.<

eagle SG

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

WoAiMeiMei SG

ok :s

add: wah cert x)

WoAiMeiMei SG

okay.. i lost some marks for the first question.. still wrong :x

im not sure about the second question

WoAiMeiMei SG

mass times friction coefficient to test slipping... or is it force. it should be mass :x ahh wth

WoAiMeiMei SG

help D:?

the force pulling the block up is the same as the one going down?

WoAiMeiMei SG

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 friction

so answer for first bit should be 281.6126N

okay second question...

WoAiMeiMei SG

for q2:

Us = g sin T / (w^2 r + g cos T)

eagle SG

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 42

Taking 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

eagle SG

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

WoAiMeiMei SG

thanks, confirms what ive done is correct