Question

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?

Answer 1

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

Answer 2

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

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Further explanation

Let's recall Moment of Force as follows:

$\boxed{\tau = F d}$

where:

τ = moment of force ( Nm )

F = magnitude of force ( N )

d = perpendicular distance between force and pivot ( m )

Let us now tackle the problem !

Given:

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

Asked:

magnitude of the force = F = ?

Solution:

If the door is in equilibrium position , 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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Answer details

Grade: High School

Subject: Physics

Chapter: Moment of Force