Question

You place a 55.0 kg box on a track that makes an angle of 28.0 degrees with the horizontal. The coefficient of static friction between the box and the inclined plane is 0.680. a) Determine the static frictional force which holds the box in place. b) You slowly raise one end of the track, slowly increasing the incline of the angle. Determine the maximum angle that the incline can make with the horizontal so that the box just remains at rest. Ms 680 u Fgsin 281 Ffg Mgm r 680 55 4 8

Answer 1

Answer:

$\theta=34 \textdegree$

Explanation:

From the question we are told that:

Mass $m=55kg$

Angle $\theta =28.0$

Coefficient of static friction $\alpha =0.680$

Generally, the equation for Newtons second Law is mathematically given by

For

$\sum_y=0$

$N=mgcos \theta$

for

$\sum_x=0$

$F_{s}=mgsin\theta$

Where

$F_{s}=\alpha*N\\\\F_{s}=\alpha*m*gcos \theta$

$F_{s}=0.68*55*9.8*cos 28$

$F_{s}=323.62N$

Therefore

$\alpha mgcos \theta=mg sin \theta$

$\theta=tan^{-1}(0.68)$

$\theta=34 \textdegree$