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The tension $ T $ at each end of a chain has magnitude 25 N (see the figure). What is the weight of the chain?

$\bullet$ The Weight of the chain: 30.1 $\mathrm{N}$

00:55

Wen Z.

04:48

Bobby B.

Calculus 3

Chapter 12

Vectors and the Geometry of Space

Section 2

Vectors

Johns Hopkins University

Campbell University

University of Michigan - Ann Arbor

University of Nottingham

Lectures

02:56

In mathematics, a vector (…

06:36

02:35

$\bullet$ Two 25.0 $\mathr…

02:00

Two 25.0-N weights are sus…

06:10

A 275 $\mathrm{N}$ bucket …

02:08

Two 25.0 -N weights are su…

0:00

In $\textbf{Fig. E5.2}$ ea…

06:17

Equilibrium of Tensions A …

14:25

Find the tension in each c…

03:37

The $10.2 \mathrm{kg}$ blo…

05:29

The ends of a chain lie in…

04:16

In Fig. E5.2 each of the s…

Talk about this question and we need to find the weight of the chain. So the weight of the chain is exactly at the center, which is downwards. And the tension is given as 25 newtons. So the tension will definitely be uh 10 ginger towards. So this is the anger. The dotted lines are actually where the tangent would be present. So if you make the components, this will be a horizontal and there will be a war. Take your component of each other attention. So the horizontal component of seacoast rita and the vertical component is going to be T 70,000 T signed 37. And this is also T signed 37 since this is at rest. So we are going to say that He signed 37 plus Pete signed 37 Is going to be equal to the weight since it is an equilibrium. So it's just a matter of substitution. So it's gonna be two t. So the two times 25 times sign off 37°. So that's going to be 50 times sine 37 As a sign. 37 times 50 is going to be a 30.09 newtons. This is the weight of the chain. Thank you.

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