A 0.60 N force is applying on a box lying on a ground the value of limiting friction is 1 N then what is the value of friction acting on the box ?
Answers
Answer:
(a) The tension in wire T attached to the hook can be determined as,
Tsin(θ)w
T=
sin(θ)
w
The force due to static friction on block A is equal to the horizontal component of the
tension.
The expression for frictional force is,
F
f
=Tcos(θ)
Substitute T=
sin(θ)
w
in the frictional force expression
F
f
=
sin(θ)
w
cos(θ)
=wcot(θ)
Substitute 12 N for w and 45
∘
for θ.
F
f
=(12N)cot(45
∘
)
=12N
Therefore friction force exerted on block A is 12N.
(b)
From the system to remain in equilibrium force of friction should be eqy=ual to the weight w of the block.
The frictional forceis,
F
f
=m
A
gμ
2
Here, m
A
is mass of A, g is acceleration due to gravity and μ
s
is coefficient
of static friction.
Substitute 60 N for mg and 0.25 for μ
s
.
F
f
=(60N)(0.25)
=15N
The force of friction should be equal to the weight w of the block so the weight of the
block is also 15 N.
Therefore, maximum weight w for the system to remain in equilibrium is 15 N.
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