Calculating Torque of a Uniform Door Due to its Own Weight

  • Thread starter Thread starter mikefitz
  • Start date Start date
AI Thread Summary
To calculate the torque of a uniform door due to its weight, the formula τ = rF is used, where τ represents torque, r is the distance from the pivot point to the line of action of the force, and F is the force applied. In this case, the door weighs 50.0 N and is 1.0 m wide, with the torque calculated about a horizontal axis at a corner. The torque is determined by multiplying the weight of the door by the distance from the pivot to the center of mass, which is located at 0.5 m from the hinge. The discussion emphasizes the importance of understanding the definition of torque and its derivation. Overall, the calculation involves applying basic physics principles to find the torque exerted by the door's weight.
mikefitz
Messages
155
Reaction score
0

Homework Statement



A uniform door
weighs 50.0 N and
is 1.0 m wide and
2.6 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?

http://img519.imageshack.us/img519/5172/untitledwp6.png


question; where does t=rF come from? it is not found on my equation sheet...
 
Last edited by a moderator:
Physics news on Phys.org
mikefitz said:
question; where does t=rF come from? it is not found on my equation sheet...

This is extremely elementary. What is the definition of the torque?
 
Thread 'Minimum mass of a block'

Thread 'Minimum mass of a block'

Here we know that if block B is going to move up or just be at the verge of moving up ##Mg \sin \theta ## will act downwards and maximum static friction will act downwards ## \mu Mg \cos \theta ## Now what im confused by is how will we know " how quickly" block B reaches its maximum static friction value without any numbers, the suggested solution says that when block A is at its maximum extension, then block B will start to move up but with a certain set of values couldn't block A reach...
View full post »
Thread 'Calculation of Tensile Forces in Piston-Type Water-Lifting Devices at Elevated Locations'

Thread 'Calculation of Tensile Forces in Piston-Type Water-Lifting Devices at Elevated Locations'

Figure 1 Overall Structure Diagram Figure 2: Top view of the piston when it is cylindrical A circular opening is created at a height of 5 meters above the water surface. Inside this opening is a sleeve-type piston with a cross-sectional area of 1 square meter. The piston is pulled to the right at a constant speed. The pulling force is（Figure 2）： F = ρshg = 1000 × 1 × 5 × 10 = 50,000 N. Figure 3: Modifying the structure to incorporate a fixed internal piston When I modify the piston...
View full post »

Similar threads

Equilibrium Problem - Balancing Torques
Replies
4
Views
3K
How Is Torque Calculated on a Hinged Door Due to Its Weight?
Replies
1
Views
5K
What is the torque on the bolt due to the worker and the weight of the beam?
Replies
1
Views
2K
How to Apply Derivatives in Physics Problems?
Replies
9
Views
1K
Torque- rigid objects in equalibrium/ center of gravity.
Replies
3
Views
5K
Static Equilibrium Door question: Calculate Horizontal Forces
Replies
31
Views
10K
Question about Normal Force and Torque
Replies
5
Views
3K
Can Statics and Torque Principles Help Solve These Physics Problems?
Replies
3
Views
3K
Is the Angular Momentum of a Pendulum Conserved?
Replies
5
Views
6K
Rotating rod of negligible mass attached to two point masses
Replies
19
Views
4K

Hot Threads

Recent Insights

Back
Top