a https/ nment FULL SCREEN Chapter 06, Problem 017 In the figure, a force Pacts
ID: 776863 • Letter: A
Question
a https/ nment FULL SCREEN Chapter 06, Problem 017 In the figure, a force Pacts on a block weighing 47.0 N. The block is initially at rest on a plane incined at angle 8-17.0° to the horizontal. The positive direction of the x axis is between block and plane are - 0.530 and -0.380. In unit-vector notation, what is the frictional force on the block from the plane when is (a) (-5.10 N) , (b) (-8.20 N) , and (c) (-15.0 N) , ? (a) Number (b) Number (c) Number Show work is REQUIRED for this question: j Units Units j Units 9pensouns Version 4.24.5 search TRAY 36Explanation / Answer
the maximum static friction force fmax = us*W*costheta = 0.53*47*cos17 = 23.8 N
the static friction force increases up to fs and if the net force on the block
is greater than fs , the bloc kmoves
component of gravitational force Fg = -W*sinthetai = -47*sin17 i = -13.74 i
(a)
net force down the incline F = P + Fg = -5.1 i - 13.4 i = -18.5 i
F < fmax
the block will not move
F + f = 0
F + fs = 0
fs = -F = 18.5 i
fs = 18.5 i + 0 j + 0 k
(b)
net force down the incline F = P + Fg = -8.2 i - 13.4 i = -21.6 i
F < fmax
the block will not move
F + f = 0
F + fs = 0
fs = -F = 21.6 i + 0j + 0 k
(c)
net force down the incline F = P + Fg = -15 i - 13.4 i = -28.4 i
F > fmax
the block will move
the kinetic friction force acts
fk = uk*W*costheta
fk = 0.38*47*cos17 = 17.1 N
fk = 17.1i + 0j +0 k
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