Recent content by Sabeshan Ratneswaran

Hi Guys,When we are finding the wasted energy when something is dropped a tower, we do this: mgH-0.5mv^2 My question is, does the velocity of the kinetic energy have to be vertical component, or can it be the impact velocity when it hits the ground? Thanks

Thank you for the consistent help.

Make sense mentor

It is not from P to F, the distant from P to Q is 2 so resolved it do 2cos(30).. then I was going to find the value of F by doing 58.86/ cos(30). But i know I have done something wrong.

I am only dealing with resolved distance components

The principle of moment is the sum of clockwise moment is equal to sum of anticlockwise moment in an equilibrium position. the expression will be FxD=FxD So basically we are dealing with perpendicular distance so the expression with be [Cos(30)x1] x [9.81x12]=[Cos(30)x2] x F So making the...

Quite confused: is this what you mean 12x9.81xcos(30)x1=102N Then divide it by cos(30)x2 to give 58.86. but we are trying to find out the force of F which holds the bonnet equilibrium. By the way the answer is not 59N.

We know the vertical component as 102 N so we can resolve to find the tension. This is where I got the expression as 102/cos(30)

Hi there, This is where I am confused. Moment about a pivot is force multiplied by the perpendicular distance between the the pivot and the line of action

Homework Statement In this I tried to resolve the components. So first thing, I converted the 12kg into Newtons so it would be 117.72 Newtons. Then found the perpendicular distance which is to g: cos(30)x1 then multiply the answer by 117.2N to give the weight down as 102N. As the moment...

Thank you mentor

Thank you indeed mentor, I got the answer (1.5V) but is there any easy way to tackle this question as it is only mark and this is only my first year in the course.

this gave 4 ohms