A chain consisting of 5 links of mass 0.1 kg each is lifted vertically upwards with a constant acceleration 5 m / s 2 as shown in figure. The force of interaction between the top link and the link immediately below it will be: ( g =10 m / s 2)

Question

A chain consisting of 5 links of mass 0.1 kg each is lifted vertically upwards with a constant acceleration 5 m / s 2 as shown in figure. The force of interaction between the top link and the link immediately below it will be: ( g =10 m / s 2) (A) 6 N (B) 4 N (C) 3 N (D) 2 N

Tardigrade Admin · Accepted Answer

F = M ( g + a )=5 × 0.1(10+5)=7.5 N for top link F-F1=m(g+a) ⇒ F 1=7.5-0.1(15)=6 N FBD of top link <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/475382ee33c3c49f9f33969a66712834-.png" />

Q. A chain consisting of $5$ links of mass $0.1 k g$ each is lifted vertically upwards with a constant acceleration $5 m / s^{2}$ as shown in figure. The force of interaction between the top link and the link immediately below it will be : $(g = 10 m / s^{2})$

6 N

4 N

3 N

2 N

$F = M (g + a) = 5 \times 0.1 (10 + 5) = 7.5 N$
for top link $F - F_{1} = m (g + a)$
$\Rightarrow F_{1} = 7.5 - 0.1 (15) = 6 N$
FBD of top link

Q. A chain consisting of 5 links of mass 0.1kg each is lifted vertically upwards with a constant acceleration 5m/s2 as shown in figure. The force of interaction between the top link and the link immediately below it will be : (g=10m/s2)

6 N

4 N

3 N

2 N

F=M(g+a)=5×0.1(10+5)=7.5N for top link F−F1​=m(g+a) ⇒F1​=7.5−0.1(15)=6N FBD of top link

Q. A chain consisting of $5$ links of mass $0.1 k g$ each is lifted vertically upwards with a constant acceleration $5 m / s^{2}$ as shown in figure. The force of interaction between the top link and the link immediately below it will be : $(g = 10 m / s^{2})$

$F = M (g + a) = 5 \times 0.1 (10 + 5) = 7.5 N$
for top link $F - F_{1} = m (g + a)$
$\Rightarrow F_{1} = 7.5 - 0.1 (15) = 6 N$
FBD of top link