



Study with the several resources on Docsity
Earn points by helping other students or get them with a premium plan
Prepare for your exams
Study with the several resources on Docsity
Earn points to download
Earn points by helping other students or get them with a premium plan
Community
Ask the community for help and clear up your study doubts
Discover the best universities in your country according to Docsity users
Free resources
Download our free guides on studying techniques, anxiety management strategies, and thesis advice from Docsity tutors
Equilibrium Notes. 1 – Translational Equilibrium. Ex. A 20.0 kg object is suspended by a rope as shown. What is the net force acting on it?
Typology: Lecture notes
1 / 6
This page cannot be seen from the preview
Don't miss anything!
Ex. A 20.0 kg object is suspended by a rope as shown. What is the net force acting on it?
Ex. Ok that was easy, now that same 20.0 kg object is lifted at a velocity of 4.9 m/s. What is the net force acting on it?
Because in both case the net force on the objects is zero they are said to be in ___________________. If the object is stationary it is said to be in ______________________________, while an object moving at a constant velocity is in _______________________________.
These are both case where the object is in ______________________ ____________________________.
Translational motion refers to motion along a line, therefore: The condition of equilibrium:
________________ = ___________________ And so,
________________ = ___________________
________________ = ___________________
Ex. A 64 N object is suspended using ropes as shown in the diagram. Calculate tensions T 1 and T 2 in the ropes.
35 o^50 o
T 1 T 2
Strategy 1: Components
Strategy 2: Create a closed vector diagram
Ex. An object is suspended as shown. If the tension in one of the ropes is 50 N as shown, what is the weight of the object?
You can use Strategy 1 or Strategy 2 , just be sure you know both ways. You’re bound to hit a brick wall eventually and it’s nice to be able to try it from a different angle, no pun intended...
37 o
Ex: Two students sit on opposite sides of an 800 N teeter-totter. Student 1 has a mass of 65 kg and sits at the very end of the teeter-totter. Student 2 has a mass of 90 kg. How far from the pivot should he sit in order to achieve equilibrium?
Extension: What are the vertical and horizontal components of the supporting force provided by the hinge in the last question?
Ex: A 3500 kg truck is parked on a bridge as shown. If the bridge deck itself has a mass of 6500 kg find the supporting force provided by each of the two support posts.
15 m
5 m
2.6 m
o
Although we’ve already learned about torque, we don’t quite have the whole story. So far we have only seen torque provided by forces acting perpendicular to the body in equilibrium. What happens if a force acts in a direction other than perpendicular to the body?
Ex A 2.2 m long 50.0 N uniform beam is attached to a wall by means of a hinge. Attached to the other end of the beam is a 100 N weight. A rope also helps support the beam as shown.
a) What is the tension in the rope? b) What are the vertical and horizontal components of the supporting force provided by the hinge?
30 o
First we draw the beam with the forces acting on it and their distances from the pivot:
So, whenever we are calculating the torque on a body we must ALWAYS use the _________________________ _______________________ of the force.
Ok now go solving! a)
b)
Notice that if we break the tension in the rope into component forces, the ______________ component does not contribute to the torque in either the clockwise or counterclockwise direction