23. A block of mass 10 kg is placed on rough inclined
plane of inclination angle 37". Coefficient of friction
between block & plane is 0.8, find minimum
magnitude of F so that block remains at rest
370
(1) 4 N
(2) 124 N
(3) 122 N
(4) Zero
(N.B- Wrong ans should be reported by brainly developers...bcz i will share screen shot)
Answers
Answer:
(4) zero is the answer of this question
Answer:
Please mark me as a brainst
Explanation:
ANSWER
Here angle of repose α=tan
−1
(μ
s
)
α=tan
−1
(
2
1
)=30
0
The angle of inclination is greater than the angle of respose. The friction force on the block will act in upward direction.
For the acceleration of block,
Mgsinθ−μN=Ma
⇒a=gsinθ−μgcosθ
=10(sin37
0
−0.5cos37
0
)
b. If external force F= 75N is applied on the block.
Let us find net driving force acting on block. Parallel to inclined two external forces are acting one in upward direction F and other is the component of weight in the direction downward the plane, Mgsinθ.
Net driving force f
driving
=F−Mgsinθ
⇒F
driving
=75−10×10×sin37
0
=75−60=15N
Maximum resisting force that oppose relative motion is maximum friction force (or f
lim
)
f
lim
=μ
s
Mgcosθ
=0.5×10×10×cos37
0
=40N
Here f
lim
<f
resisting
. Hence, the block will not move and friction will be static will act in the direction opposite to driving force, i.e., in downward direction.
c. To move block, the least value of driving force should be 40 N. But in above case, driving force is 15N (up). Hence, if we add △F=25N in upward direction, the block will overcome maximum resistence force (or friction) and starts moving up.
∴60+40=75+△F⇒△F=25N
d. As resisting force which is maximum friction force is 40N and the component of weight parallel to incline is 60 N and acting downward. If we remove F, then the driving forces will be the only component of the weight in the direction downward the incline plane. In this case, friction will act in upward direction. Hence, the required value of F to make block in equilibrium,
F + 40 = 60
or F = 20N.
