Posted by **Julia** on Wednesday, April 27, 2011 at 12:19pm.

A trap door, of length and width 1.65 m, is held open at an angle of 65.0 degrees with respect to the floor. A rope is attached to the raised edge of the door and fastened to the wall behind the door in such a position that the rope pulls perpendicularly to the trap door. If the mass of the trap door is 16.8 kg, what is the torque exerted on the trap door by the rope?

I've seen several incorrect answers to this exact question when I searched online. I thought I would provide the correct general method on how to do this problem and give you the answer you should get if it is done correctly. (see below for method)

- Physics -
**Julia**, Wednesday, April 27, 2011 at 12:33pm
All right. Think of the trap door as a bar and draw a free-body diagram (FBD) with it as such. Assuming this is a uniform system, the center of mass will also be at the center of the door and that is also where gravity will be pulling on the door. Since the door is cocked at an angle to the floor, the pull of gravity will also be at an angle. Calculate this and make sure you draw the gravity in the correct orientation in your FBD. The force of the rope is obviously perpendicular to the door, so keep that in mind.

Now, the system is in equilibrium, therefore, the sum of the torques due to gravity (Tg) and the torque due to the rope (Tc) must be equal to zero.

Tc-Tg=0

Tc=Tg

Now, you have to find the force of the rope pulling on the door from this equation, using the fact that you know that T=rFsin(x).

Tc=rFsin(90)

where r is radius from hinge to rope, F is force of rope pulling on door, and 90 is the angle of force to door

Tg=rFsin(x)

where r is radius from hinge to center mass, F is force due to gravity, and x is angle at which you calculated gravity was pulling on the door.

rFsin(90)=rFsin(x)

Rearrange to get force due to rope

F=rFsin(x)/rsin(90)

This gives you the force with which the rope pulls on the door.

Now, you have a force you can plug into your torque equation T=rFsin(x). Plug it in and solve for torque.

For this particular problem using these values, you should get that the torque exerted on the trap door is equal to 57.4 N*m. I hope this helps any of you who might be struggling with this problem.

## Answer This Question

## Related Questions

- Physics - trap door of length and width 1.65 m is held open at an angle of 65.0 ...
- Physics - trap door of length and width 1.65 m is held open at an angle of 65.0 ...
- Physics - A 46.4-N force is applied to the outer edge of a door of width 1.26 m ...
- PHYSICS! - A horizontal force of 10 N is applied to the outer edge of a door at...
- physics - A small boy is trying to open a 50 kg door with a height of 2 meters ...
- physics - A 51.4 N force is applied to the outer edge of a door of width d = 2....
- Physics - A 15 kg, 1m wide door which has frictionless hinges is closed but ...
- Physics - A 15 kg, 1m wide door which has frictionless hinges is closed but ...
- Physics - A 15 kg, 1m wide door which has frictionless hinges is closed but ...
- heeeeeeeeeeeeeelp Physics - A 15 kg, 1m wide door which has frictionless hinges ...

More Related Questions