Recent content by Ashu2912

Correct, since the horizontal acceleration is zero (system is in equilibrium), the (vector) sum of forces resolved along the horizontal must be zero. You have T1cosθ towards left and T2cosθ towards right. Now, if a body is in equilibrium, it's acceleration is zero implies the vector sum of the...

The first equation is not. When you resolve T1 and T2 in the horizontal direction, you get T1 cosθ and T2cosθ, but in which directions?

What made you think that? No, you have three forces T1 ,T2 and the weight mg. Resolve these in the vertical and horizontal directions and you will get two equations. What I meant here was that (1) The equations you get are independent (2) It is not always required to resolve in two mutually...

No, they both are independent equations. You have to resolve(split vectorially) the forces in the "direction along which the acceleration component is known", which here is all possible directions, as the object is in equilibrium. These directions may be mutually perpendicular but it is not...

Homework Statement This is a diagram from "Structural Analysis" by Aslam Kassimalli while solving frames using slope-deflection method. In the figure, the tangent to the deflected shape at C` is parallel to the original member AC implying that the slope at C is zero. However, later they have...

I'm not referring to the unit (pounds). As BvU said, the unit of the scale in in N (Newtons). I mean what quantity does it measure? Draw the free body diagram of the man. The forces acting on him in a stationary elevator will be (1) His weight and (2) Normal reaction from the weighing machine...

First of all, the direction of the normal force Fn is incorrect. It should always be "perpendicular to the contact surface".

What does a weighing scale usually read, the weight of something else actually? Know that and you will get your answer! :wink:

You are given the value of the force F from t=0 to t=6. So, you can find the acceleration for t=0 to t=3, then t=3 to t=5, then t=5 to t=6. Using the equations of motion in the x-direction, you can find the velocity at the end of t=3 (which will be the velocity at the beginning of t=3 to t=5)...

This is a diagram I found in "Structural Analysis" by Aslam Kassimalli. Here, if the member AC is inextensible, shouldn't the point c`in the deflected frame lie on the line CD in the original frame?

In analysis of frames, we say that "The members of the frames are axially inextensible". In context of frames with vertical legs, what I understand is that the length of the chord joining the member ends in the displaced condition would be same in the axial direction. However, what does it imply...

Using conservation of energy, the energy of roller coaster+earth system is conserved, in an ideal case, i.e. change in potential energy = change in kinetic energy. However, in this case there is 50% loss of energy, i.e. only half of the potential energy change equals the kinetic energy change. I...

Sorry I mean infinity to zero! And what doc al said is absolutely correct!

Hey dreamlord, I think I have understood this now! Fext(vector) =- qE(vector) like you said. However, while integrating you have to consider dr(vector) only, the integration limits being from 0 to infinity!

It means that the collision between the particle and the surface is perfectly inelastic.