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9 days 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:

Ostrovityanka [942]9 days ago

8 0

The greatest force magnitude, F, that allows the door to stay stationary is approximately 265 N

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Additional information

Let’s revisit the concept ofMoment of Force in this manner:

$\boxed{\tau = F d}$

where:

τ = moment of force ( Nm )

F = magnitude of force ( N )

d = perpendicular distance from force to pivot ( m )

Now, let’s solve the issue!

Given:

door's weight = w = 145 N

force direction = θ = 20°

distance from hinge to force application = d = 2.50 m

door length = L = 3.13 m

To find:

force magnitude = F =?

Solution:

If the door is in an equilibrium state, then:

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

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

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

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

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Learn more

Gravity Impacts:
Gravity's Effect on Objects:
The Acceleration Due to Gravity:
Newton's Laws of Motion:
Newton's Law Example:

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Answer context

Grade: High School

Subject: Physics

Chapter: Moment of Force

serg [1.1K]9 days ago

4 0

<span>Conclusion: The door's weight results in a CCW torque which can be calculated as Tccw = 145 N*3.13 m / 2. You require a CW torque that balances this Tcw = F*2.5 m*sin20</span>

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12 hours ago

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ValentinkaMS [1149]

Answer:

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Response:

3.5 N

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6 days ago

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The scenario is as follows:

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→ K.E = 200 m joules when the ball strikes the ground.

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