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monitta
3 years ago
12

A one-piece cylinder has a core section protruding from the larger drum and is free to rotate around its central axis. A rope wr

apped around the drum of radius 1.22 m exerts a force of 3.11 N to the right on the cylinder. A rope wrapped around the core of radius 0.62 m exerts a force of 6.92 N downward on the cylinder. 3.11 N 6.92 N What is the magnitude of the net torque acting on the cylinder about the rotation axis? Answer in units of N · m.
Physics
2 answers:
Rus_ich [418]3 years ago
7 0

Answer:

8.0846 N-m

Explanation:

The torques are in the same ( clockwise ) direction and so all we need to do is add the by calculating the torque and add

1.22×3.11 + 0.62×6.92 = 8.0846 N-m

magnitude of the net torque acting on the cylinder about the rotation axis

= 8.0846 N-m

11Alexandr11 [23.1K]3 years ago
6 0

Answer:

The net torque is 0.4962 N m

Explanation:

please look at the solution in the attached Word file

Download docx
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3 years ago
A uniform metal tube of length 5m and mass 9kg is suspended by two vertical wires attached at 50cm and 150cm respectively from t
Elena-2011 [213]

Answer:

force (tension) of 29.4 N (upward)  in 100 cm

force (tension) of 58.4 N (upward)  in 200 cm

Explanation:

Given:

Length of tube = 5 m (500 cm)

Mass of tube = 9

Suspended vertically from 150 cm and 50 cm.

Computation:

Force = Mass × gravity acceleration.

Force = 9.8 x 9

Force = 88.2 N

So,

Upward forces = Downward forces

D1 = 150 - 50 = 100 cm

D2 = 150 + 50 = 200 cm

And F1 = F2

F1 x D1 = F2 x D2

F1 x 100 = F2 x 200

F = 2F

Total force = Upward forces + Downward forces

3F = 88.2

F = 29.4 and 2F = 58.8 N

force (tension) of 29.4 N (upward)  in 100 cm

force (tension) of 58.4 N (upward)  in 200 cm

4 0
3 years ago
The distance that a spring will stretch varies directly as the force applied to the spring. A force of 8080 pounds is needed to
xxTIMURxx [149]

Answer:

F₂= 210 pounds

Explanation:

Conceptual analysis

Hooke's law

Hooke's law establishes that the elongation (x) of a spring is directly proportional to the magnitude of force (F) applied to it, provided that said spring is not permanently deformed:

F= K*x   Formula (1)

Where;

F  is the magnitude of the force applied to the spring in Newtons (Pounds)

K is the elastic spring constant, which relates force and elongation. The higher its value, the more work it will cost to stretch the spring. (Pounds/inch)

x the elongation of the spring (inch)

Data

The data given is incorrect because if we apply them the answer would be illogical.

The correct data are as follows:

F₁ =80 pounds

x₁= 8 inches

x₂= 21  inches

Problem development

We replace data in formula 1 to calculate  K :

F₁= K*x₁

K=( F₁) / (x₁)

K=( 80) / (8) = 10 pounds/ inche

We apply The formula 1 to calculate  F₂

F₂= K*x₂

F₂= (10)*(21)

F₂= 210 pounds

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Answer:

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