1) An inclined plane with angle ?=30.9 degrees has a rough surface with coeffici
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Question
1) An inclined plane with angle ?=30.9 degrees has a rough surface with coefficient of static friction ?s=0.56 and kinetic friction ?k=0.21. A mass, m=10.1 kg, sits at rest with respect to the plane and is attached by a string strung over a pulley to a hanging mass, M.
What is the maximum value of the hanging mass in kg, Mmax, such that the mass, m, remains stationary on the inclined plane?
and
2) An inclined plane with angle ?=47 degrees has a rough surface with coefficient of static friction ?s=0.88 and kinetic friction ?k. A mass, m=11.1 kg, sits at rest with respect to the plane and is attached by a string strung over a pulley to a hanging mass, M, such that the mass, m, remains stationary on the inclined plane.
The string is suddenly cut and the mass m begins to slide down the plane with an acceleration, a=3.97 m/s2. What is the value of coefficient of kinetic friction, ?k ? (g=9.80 m/s2)
Both questions use this picture, thanks
Explanation / Answer
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Que. 1,An inclined plane with angle ?=45.8 degrees has a rough surface with coefficient of static friction ?s=0.52 and kinetic friction ?k=0.2. A mass, m=11.8 kg, sits at rest with respect to the plane and is attached by a string strung over a pulley to a hanging mass, M.
Which of the following is the maximum value of the hanging mass in kg, Mmax, such that the mass, m, remains stationary on the inclined plane?
2.An inclined plane with angle ?=37 degrees has a rough surface with coefficient of static friction ?s=0.8 and kinetic friction ?k. A mass, m=9.7 kg, sits at rest with respect to the plane and is attached by a string strung over a pulley to a hanging mass, M, such that the mass, m, remains stationary on the inclined plane.
The string is suddenly cut and the mass m begins to slide down the plane with an acceleration, a=3.86 m/s2. What is the value of coefficient of kinetic friction, ?k ? (g=9.80 m/s2)
Ans.
1. when the system remains stationary,
accelaration = 0
therefore, for maximum value M, the friction force is acting down the plane because the block will try to go up the plane
Mg - mgsin? - ?mgcos? = 0
M = (m)(sin? - ?cos?)
M= 11.8(sin45.8 - 0.52cos45.8)
M = 4.1817 kg
2.
mg sin? - ?k*mg*cos? = ma
?=37
g=9.8
a=3.86
solving we have
?k = 0.26
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