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irinina [24]
2 years ago
13

Two objects are placed on a scale so that it balances. One object weighs 5 N and is placed 0.5 m from the fulcrum of the scale.

The other object is placed 1 m from the fulcrum. How heavy is the other object?
Physics
1 answer:
insens350 [35]2 years ago
5 0

The weight of the other object placed 1 m from the fulcrum that balances the other object is 2.5 N.

<h3>Principle of moment</h3>

The weight of the second object is determined by applying principle of moment as shown below;

take moment about the pivot,

sum of clockwise moment = sum of anticlockwise moment

F₁r₁ = F₂r₂

5 x 0.5 = F₂ x 1

2.5 N = F₂

Thus, the weight of the other object placed 1 m from the fulcrum that balances the other object is 2.5 N.

Learn more about moment here: brainly.com/question/6278006

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When two oppositely charged electrodes are immersed in a solution, positively charged ions are attracted to the negative electro
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A prismatic bar AB of length L and solid circular cross section (diameter d) is loaded by a distributed torque of constant inten
Lyrx [107]

Answer:

a) the maximum shear stress τ_{max} the bar is 16T_{max} /πd³

b) the angle of twist between the ends of the bar is 16tL² / πGd⁴  

Explanation:

Given the data in the question, as illustrated in the image below;

d is the diameter of the prismatic bar of length AB

t is the intensity of distributed torque

(a) Determine the maximum shear stress tmax in the bar

Maximum Applied torque  T_max = tL

we know that;

shear stress τ = 16T/πd³

where d is the diameter

so

τ_{max} = 16T_{max} /πd³

Therefore, the maximum shear stress τ_{max} the bar is 16T_{max} /πd³

(b) Determine the angle of twist between the ends of the bar.

let theta (\theta) be the angle of twist

polar moment of inertia I_p} = πd⁴/32

now from the second image;

lets length dx which is at distance of "x" from "B"

Torque distance x

T(x) = tx

Elemental angle twist = d\theta = T(x)dx / GI_{p}

so

d\theta = tx.dx / G(πd⁴/32)

d\theta = 32tx.dx / πGd⁴

so total angle of twist \theta will be;

\theta =  \int\limits^L_0  \, d\theta

\theta =  \int\limits^L_0  \, 32tx.dx / πGd⁴

\theta = 32t / πGd⁴  \int\limits^L_0  \, xdx

\theta = 32t / πGd⁴ [ L²/2]

\theta = 16tL² / πGd⁴  

Therefore,  the angle of twist between the ends of the bar is 16tL² / πGd⁴  

7 0
3 years ago
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