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Brilliant_brown [7]

2 years ago

9

One end of a rope is tied to the handle of a horizontally-oriented and uniform door. a force fis applied to the other end of the

rope as shown in the drawing. the door has a weight of 145 n and is hinged on the right. what is the maximum magnitude of ffor which the door will remain at rest?

2 answers:

olya-2409 [2.1K]2 years ago

8 0

The maximum magnitude of F for which the door will remain at rest is about 265 N

$\texttt{ }$

<h3>Further explanation</h3>

<em>Let's recall </em><em>Moment of Force</em><em> as follows:</em>

$\boxed{\tau = F d}$

<em>where:</em>

<em>τ = moment of force ( Nm )</em>

<em>F = magnitude of force ( N )</em>

<em>d = perpendicular distance between force and pivot ( m )</em>

Let us now tackle the problem !

<u>Given:</u>

weight of the door = w = 145 N

direction of the force = θ = 20°

distance between hinge and the applied force = d = 2.50 m

length of the door = L = 3.13 m

<u>Asked:</u>

magnitude of the force = F = ?

<u>Solution:</u>

<em>If the door is in equilibrium position , then :</em>

$\texttt{Total Clockwise Moment at Hinge = Total Anticlockwise Moment at Hinge }$

$F \times d \times \sin \theta = w \times \frac{1}{2} L$

$F \times 2.50 \times \sin 20^o = 145 \times \frac{1}{2} (3.13)$

$\boxed {F \approx 265 \texttt{ N}}$

$\texttt{ }$

<h3>Learn more</h3>

Impacts of Gravity : brainly.com/question/5330244
Effect of Earth’s Gravity on Objects : brainly.com/question/8844454
The Acceleration Due To Gravity : brainly.com/question/4189441
Newton's Law of Motion: brainly.com/question/10431582
Example of Newton's Law: brainly.com/question/498822

$\texttt{ }$

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Moment of Force

sergeinik [125]2 years ago

4 0

<span>Answer:The weight of the door creates a CCW torque given by Tccw = 145 N*3.13 m / 2 You need a CW torque that's equal to that Tcw = F*2.5 m*sin20</span>

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Answer:

Second object is located at 42.03 cm in front of mirror

Explanation:

In this question we have given,

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we have to find the distance of second object from convex mirror

we know that u, v and f are related by following formula

$\frac{1}{f} =\frac{1}{v}+ \frac{1}{u}$ .............(1)

put values of u and v in equation (1)

we got,

$\frac{1}{f} =\frac{1}{7.05}+ \frac{1}{-13.5}$

$\frac{1}{f}=\frac{13.5-7.05}{13.5\times 7.05}$

$\frac{1}{f}=\frac{6.45}{13.5\times 7.05}\\f=13.5\times 1.09\\f=14.75$

we have given that

second object is twice as tall as the first object

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it means

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we know that

$\frac{v}{u}=\frac{i}{o}\\i=\frac{o\times v}{u}$

therefore,

$i_{1}=\frac{o_{1}\times v}{u}$ .................(3)

put values of v and u in equation 3

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$i_{1}=-0.52o_{1}$

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we know that

$i_{2}=\frac{o_{2}\times V}{U}$ .................(4)

put value of $i_{2}$ and $o_{2}$ in equation 4

$-.52o_{1}=\frac{2o_{1}\times V}{U}$

$U=\frac{2o_{1}\times V}{-.52o_{1}} \\U=-3.85V$

we know that U,V and f are related by following formula

$\frac{1}{f} =\frac{1}{V}+ \frac{1}{U}$ .............(5)

put values of f and U in equation 5

we got

$\frac{1}{14.75} =\frac{1}{V}- \frac{1}{3.85V}$

$\frac{1}{14.75} =\frac{2.85}{3.85V}$

$\frac{1}{14.75} =\frac{2.85}{3.85V}\\V=\frac{2.85\times 14.75}{3.85}\\V=10.91 cm$

Therefore,

U=-10.91\times 3.85

U=-42.03 cm

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