wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

Three blocks of mass 10 kg, 5 kg and 8 kg are connected to each other with a light string and placed on a wedge with angle of inclination 30. A force F applied on the string causes all the three blocks to move with a common acceleration of 2 m/s2. If the coefficient of friction between the blocks and the surface of the wedge is 0.2, then what is the tension at the three points A,B and C?


A
TA=200.8 N, TB=69.85 N, TC=113.51 N
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
TA=113.51 N, TB=69.85 N, TC=200.8 N
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
TA=200.8 N, TB=113.51 N, TC=69.85 N
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
TA=195 N, TB=60 N, TC=100 N
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is C TA=200.8 N, TB=113.51 N, TC=69.85 N
FBD of the three blocks taken as a system:



FBD of block C:



FBD of block B and C taken as a system:


The entire system accelerates at the rate of 2 m/s2. Considering all the three masses together, we get,

Fμ(m1+m2+m3)gcosθ(m1+m2+m3)gsinθ =(m1+m2+m3)aF0.2×(10+5+8)×10×cos30(10+5+8)×10×sin30 =(10+5+8)×2F=200.8 N
Therefore tension at point A :TA=200.8 N

Due to the force exerted at point B, the other two blocks move with an acceleration of 2 m/s2.
From the free body diagram, we get,
FB0.2(5+8)×10×cos30(5+8)×10×sin30 =(5+8)×2FB=113.51 N
Therefore tension at point B :TB=113.51 N

Due to the force exerted at point C, the last block moves with an acceleration of 2 m/s2

From the free body diagram, we get,
FC0.2×8×10×cos308×10×sin30 =8×2FC=69.85 N
Therefore tension at point C :TC=69.85 N

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