Recent content by thermodynamicaldude

I'm not sure, but if you say that the tension of the dashpot is Z_d*(x_1 - x_2) , then this is the magnitude of the force the dashpot exerts on the strings. Using Newtons third law, the strings will exert an opposite and equal force on the dashpot. Thus, we would have Z_d*(x_1 - x_2) =...

I am currently trying out for the US Physics team, and am working on some practice exams. I was wondering if anyone could offer some assistance on the following problems below: Problem A2 http://www.compadre.org/psrc/evals/IPO_Exam_2_2003.pdf Problem's A4 and B1...

The wording in your description may be slightly confusing to posters. Perhaps if you could include a diagram of the problem. Anyways...when you work with statics...always remember that about any point, the torques always = 0. Also, sum of the x forces = 0 and sum of the y forces = 0.

Net work done on object = (horizontal component of 118 - frictional force) * distance Net work done on object = change in kinetic energy (work-energy theorem)

The rope must also overcome the force of gravity acting on Tarzan when he is at the bottom of the arc. Thusly, the tension of the rope = Fcent + Fgrav at this point.

The question is asking for "when", which implies time. THus, take the derivatives, and solve for t when the derivative = 0. AFter you have found your values for t, then you will need to test whether the value is a maximum or a minimum. (if you don't find a max or min. with these values...

...x = v cos (theta) t divide both sides by vcos(theta)...and you get t = x/(v cos (theta) Now plug this into the equation for y... When you do this, you should be able to solve for initial velocity.

To start off... Gravitational Potential Energy = Linear Kinetic Energy + Rotational Kinetic Energy ...Rotational Kinetic Energy = 1/2 I w^2 w(its called omega) = v/r I of Sphere = (2/5)mr^2... ..thusly...Rotational Kinetic Energy = (1/2)[(2/5)mr^2][v/r]^2...which simplifies to...

Tarzan is moving in an arc...that means that centripetal acceleration is involved...(since he's moving in a circular direction). Thusly, there is a centripetal force acting on Tarzan (Fc = mv^2/R) and the graviational force as well (F = mg)...basically...work with that info...and also...try...

P = F/A; Change in momentum = Impulse = Force * time Change in energy = Work = (Force)(distance)(cos (theta))...since this a rocket..the force will most likely be in the same direction as the movement...so...cos(0) = 1...and you have... Work = Force(distance). Or, since you know the...

...think...when the car is at rest...what forces are acting on it? What forces are acting on the ball? Is the ball in equilibrium? ...when the car is accelerated up the incline...how does this acceleration affect the ball (remember, princaple of equivalence?)

The force a spring exerts on an object is F = -kx.

yeah...sorry 'bout that..I left out the sumation part... but basically follow Spectre5's advice...free body diagrams are essential to solving most force related Physics problems

F = ma Thus, when the car is accelerating, the force acting on the car not only must overcome the force of gravity, but it also must accelerate the car.

Looks like you're using mr^2 for the moment of Inertia...which would work if it were a ring.. ..however, the moment of Inertia of a thick solid disk is: I = 0.5mr^2.. ...hope that helps!