(0%) Problem 9: A block of mass m1 = 1.5 kg rests on an inclined plane with a co
ID: 1575869 • Letter: #
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(0%) Problem 9: A block of mass m1 = 1.5 kg rests on an inclined plane with a coefficient of static friction of u0.105 between the block and the plane. The inclined plane is L = 6.6 m long and it has a height of 3.6 m at its tallest point t- ©theexpertta.com 20% Part (a) what angle, in degrees, does the plane make with respect to the horizontal? Grade Summary 0% 100% Potential cosO cotan0 asin acos0 Submissions Attempts remaining: 10 (0% per attempt) detailed view sin 78 9 atan() acotanO s acotanO S sinhO cosh0tan0 coho Degrees O Radians 0 END DEL CLEAR Submit Hint I give up! Hints: 2% deduction per hint. Hi ints remaining: 2 Feedback: 1% deduction per feedback. 20% Part (b) What is the magnitude of the normal force, FN in newtons, that acts on the block? 20% Part (c) What is the component of the force of gravity along the plane, Fgx in newtons? - 20% Part (d) Write an expression, in terms of , the mass m, the coefficient of static friction , and the gravitational constant g, for the magnitude of the force due to static friction, Fs, just before the block begins to slide - 20% Part (e) Will the block slide?Explanation / Answer
a) From geometry
= arcsin(h/L) = arcsin(3.6/6.6)=33.06°
b) Fn = m*g*cos =11.5*9.8*cos33.06= 94.44 N
c) Fgx = m*g*sin =11.5*9.8*sin33.06 =61.48 N
d) Fs = µ*m*g*cos = 0.105*11.5*9.8*cos33.06=9.916 N
e) Yes; Fgx > Fs
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