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A 10 kg box is at rest at the bottom of a ramp with a 35" inclineThe box is attached by a rope to a 35 kg weight hanging off a pulleywith negligible friction at the top of the ramp. What is the minimumcoefficient of friction needed between the box and the ramp for thebox to remain at rest?35.0

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A 10 kg box is at rest at the bottom of a ramp with a 35" incline
The box is attached by a rope to a 35 kg weight hanging off a pulley
with negligible friction at the top of the ramp. What is the minimum
coefficient of friction needed between the box and the ramp for the
box to remain at rest?
35.0
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A 10 kg box is at rest at the bottom of a ramp with a 35" incline The box is attached by a rope to a 35 kg weight hanging off a pulley with negligible friction at the top of the ramp. What is the minimum coefficient of friction needed between the box and the ramp for the box to remain at rest? 35.0

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Step 1

Given:

Mass of the box, m1 = 10 kg

Mass of the rope, m2 = 35 kg

Let, the coefficient of friction = µk

Step 2

For inclined plane, equation of motion can be given as

т
та %3Dтg sin @ — д;R-T
T
R
T m2g
та 3 т,g sin @ — MaR-m,g
for rest
35.0
та %3D 0
migsine
m2g
mgsin 0- R-m2g = 0
for equilibrium
migcose
тig
R m,gcose
therefore
тgsin @ - M; m,gcos@ — m,g %3D0
mgsin0-mgcos0 = m2g
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т та %3Dтg sin @ — д;R-T T R T m2g та 3 т,g sin @ — MaR-m,g for rest 35.0 та %3D 0 migsine m2g mgsin 0- R-m2g = 0 for equilibrium migcose тig R m,gcose therefore тgsin @ - M; m,gcos@ — m,g %3D0 mgsin0-mgcos0 = m2g

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Step 3

On putting the...

10 x9.8x sin 35° - /i, x10x9.8x cos 35° = 35x9.8
98 x0.574-x 98 x0.819 = 343
343-56.252
80.262
286.748
80.262
=3.573
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10 x9.8x sin 35° - /i, x10x9.8x cos 35° = 35x9.8 98 x0.574-x 98 x0.819 = 343 343-56.252 80.262 286.748 80.262 =3.573

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