CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

A block of mass m on a rough horizontal surface is acted upon by two forces as shown in figure. For equilibrium of block, the coefficient of friction between block and surface is
695208_043f25a4b8df443ea39a30440e0420d0.png

A
F1+F2sinθmg+F2cosθ
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
F1cosθ+F2mg+F2sinθ
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
F1+F2cosθmg+F2sinθ
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
F1+F2cosθmg+F2cosθ
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is D F1+F2sinθmg+F2cosθ
Normal force acting on the block R=mg+F2cosθ,
Friction force f=μR
f=μ[mg+F2cosθ]
Also force is balanced in horizontal direction, f=F1+F2sinθ
Equating,
μ[mg+F2cosθ] =F1+F2sinθ
We get μ=F1+F2sinθmg+F2cosθ.
723158_695208_ans_d6a9d0c692414a658729b8e6bdd691a7.png

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
So You Think You Know Work?
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon