Question icon
Grade upto college level Mechanics

A chain consisting of five links, each with mass 100 g, is lifted vertically with a constant acceleration of 2.50 m/s2, as shown in Fig. 3-35. Find (a) the forces acting between adjacent links, (b) the force F exerted on the top link by the agent lifting the chain, and (c) the net force on each link.
src=data:image/png;base64,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

Profile image of Amit Saxena
11 Years agoGrade upto college level
Answers icon

2 Answers

Profile image of Navjyot Kalra
11 Years ago
232-1617_1.JPG
232-765_1.JPG
232-673_1.JPG
Profile image of Kushagra Madhukar
5 Years ago
Dear student,
The answer to your question will be
 
a)On each link the force from
Upward link = 1.23, 2.46, 3.69, 4.92 respectively from bottom to top (in N)
Downward link = 0, 1.23, 2.46, 3.69 respectively from bottom to top ( in N)
 
b)6.15 N
 
c)0.25 N
For complete solution refer to the above answer
 
Thanks and regards,
Kushagra