Chapter 06, Problem 017 In the figure, a force acts on a block weighing 49.0 N.

Question

Chapter 06, Problem 017 In the figure, a force acts on a block weighing 49.0 N. The block is initially at rest on a plane inclined at angle = 19.0° to the horizontal. The positive direction of the x axis is up the plane. The coefficients of friction between block and plane are 0.530 and Pk-0.350. In unit-vector notation, what is the frictional force on the block from the plane when P is (a) (-5.20 N) , (b) (-8.20 N) ,and (e) (-15.0 N) 1? (a) Number (b) Number (c) Number Click if you would like to Show Work for this question: jUnits j Units j Units Open Show Work

Explanation / Answer

Given,

Force, P = 49 N

Resolving the force in horizontal and vertical components, so

Block's normal force, Fn = 49 x cos19° = 46.33 N

downslope weight, Fg = 49 x sin19° = 15.95 N

static friction, Fs = µ x Fn = 0.53 x 46.33 = 24.56 N

and kinetic friction, Fk = 0.35 x 46.33 = 16.22 N

To overcome the static friction, you need an applied force, P = 24.56 - 16.22 = 8.34 N

The first two cases don't meet that threshold.

(a) friction, F = (15.95 + 5.20) i = 21.15 N i

(b) friction, F = 15.95 + 8.20 = 24.15 N I

(c) friction, F = 16.22 N i

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