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

Mass of block A is 100 kg and that of block B is 200 kg. As shown in figure, the block A is attached to a string tied to the wall. The coefficient of static friction between blocks A and B is (μs)A=0.2 and the coefficient of static friction between block B and floor is (μs)B=0.3. Then calculate the minimum value of force F required to move the block B. (g=10 m/s2)


A
980 N
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
1000 N
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
1100 N
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
1200 N
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is C 1100 N

From the FBD of A:
T=f1........(1)
From FBD of B:
F=f1+ f2......(2)

Since we want the minimum value of F (for just slipping condition), we will take the limiting values of friction.
f1=(μs)A mA g .........(3)
and f2=(μs)B(mA+mB)g ......(4)

Substituting (3) and (4) in (2),
F=f1+f2
=(0.2)(100g)+(0.3)(300g)=110 g=1100 N

flag
Suggest Corrections
thumbs-up
0
similar_icon
Similar questions
View More
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon