Question #153194

A 5kg block was placed on an inclined plane(50 degrees with respect to the vertical).Forces A is at angle C with respect to the horizontal. Force B is parallel with the incline.The coefficients of static and kenetic friction are 0.3 and 0.2 respectively.

A=197KN

B=49 KN

C=7 degrees

A.DETERMINE IF THERE IS MOTION AND ITS DIRECTION

B.BASED ON YOUR ANSWER IN "A" WHAT IS THE ACCELERATION OF THE BLOCK?


1
Expert's answer
2020-12-31T08:07:02-0500

A.


N=mgcos40°+197000sin47°=59.8cos40°+197000sin47°=N=mg\cos40°+197000\cdot\sin47°=5\cdot9.8\cdot\cos40°+197000\cdot\sin47°=


=144114(N)=144114(N)


49000+mgsin40°197000cos47°=49000+59.8sin40°197000cos47°=49000+mg\sin40°-197000\cos47°=49000+5\cdot9.8\cdot\sin40°-197000\cos47°=


=85322(N)=-85322(N)


Ff=0.3N=0.3144114=43234(N)F_f=0.3N=0.3\cdot144114=43234(N)


Ftotal=85322+43234=42088(N)F_{total}=-85322+43234=-42088(N)


The block will move up the inclined plane. Answer


B.


a=F/m=42088/5=8418(m/s2)a=F/m=42088/5=8418(m/s^2) . Answer











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