Magnitude of torque on a door

[trainumc]

Homework Statement

A uniform door weighs 53.0 N and is 0.9 m wide and 2.5 m high. What is the magnitude of the torque due to the door's own weight about a horizontal axis perpendicular to the door and passing through a corner?

Homework Equations

torque is equal to the perpendicular component of the force with the shortest distance between the roation axis and the point of application of the force

and

τ=r(Fsinθ)

The Attempt at a Solution

530*0.9*2.5*(sin90)


i really don't know where to begin, my teacher hasn't taught this yet, even though the webassign is due tomorrow. am i using the wrong formulas?

 

[LowlyPion]

Homework Statement

A uniform door weighs 53.0 N and is 0.9 m wide and 2.5 m high. What is the magnitude of the torque due to the door's own weight about a horizontal axis perpendicular to the door and passing through a corner?

Speaking generally you need to figure the distance of the Center of Mass as the point that you would have its weight act through relative to the desired axis of rotation.

 

[thomate1]

I would first clarify the given information and assumptions. Is the door assumed to be a rectangular shape? Are we assuming that the door is attached to a hinge at the corner where the torque is being calculated? Are we assuming that the door is in a static position and not moving? These details are important in determining the appropriate equations and solutions.

Assuming that the door is a uniform rectangular shape and is attached to a hinge at the corner where the torque is being calculated, we can use the equation τ=r(Fsinθ) to calculate the torque. In this case, the force (F) is the weight of the door (53.0 N) and the distance (r) is the shortest distance between the rotation axis (the hinge) and the point of application of the force (the center of mass of the door). Since the door is 0.9 m wide and 2.5 m high, the center of mass would be located at (0.45 m, 1.25 m) from the hinge. The angle θ would be 90 degrees, as the force is acting perpendicular to the rotation axis.

Plugging in these values into the equation, we get:

τ = (0.45 m)(53.0 N)(sin90) = 23.85 N∙m

Therefore, the magnitude of the torque due to the door's own weight about a horizontal axis perpendicular to the door and passing through a corner is 23.85 N∙m.