Two blocks of masses
m = 5kg and m, = 6kg are
connected by a light string
passing over a light
frictionless pulley as shown.
The mass m, is at rest on the
inclined plane and mass m2
hangs vertically. If the angle
of incline (0) = 30', what is
magnitude and direction of
force of friction on the 5kg
block (g = 10ms 2)
Answers
Answered by
0
Answer
Explanation:
Force along plane, F
1
=mgsinθ=25N
Balancing forces along the plane;
∴T−mgsin30
0
−f=0
f=T−mgsin30
0
=60−25
=35 N
Answered by
0
Answer:
For equilibrium condition
m2g= m1gsin30+f
60= 50/2 + f
f= 35N
friction force will act downward direction.
Hope it will help you
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