Questions: One end of a rope is tied to the handle of a horizontally-oriented and uniform door. A force F is 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 F for which the door will remain at rest? 145 N 265 N 381 N 424 N 530 N

One end of a rope is tied to the handle of a horizontally-oriented and uniform door. A force F is 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 F for which the door will remain at rest?
145 N
265 N
381 N
424 N
530 N

Transcript text: One end of a rope is tied to the handle of a horizontally-oriented and uniform door. A force $\vec{F}$ is 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 $\vec{F}$ for which the door will remain at rest? 145 N 265 N 381 N 424 N 530 N

Solution

Solution Steps

Step 1: Setting up the torque balance equation

For the door to remain at rest, the net torque acting on it must be zero. The torque due to the weight of the door acts clockwise, while the torque due to the force F acts counter-clockwise. The torque due to the weight is given by the weight multiplied by the horizontal distance from the hinge to the center of gravity of the door (half the door's width). The torque due to the force F is given by the vertical component of F (Fsin20) multiplied by the horizontal distance from the hinge to the handle.

Step 2: Calculating the torques

Torque due to weight:

τ_weight = (145 N) * (3.13 m / 2) = 227.725 Nm

Torque due to force F:

τ_F = F * sin(20°) * 2.50 m

Step 3: Solving for F

Setting the torques equal to each other:

227.725 Nm = F * sin(20°) * 2.50 m

Solving for F:

F = 227.725 Nm / (sin(20°) * 2.50 m) ≈ 266.7 N