Recent content by DumSpiroSpero

Thanks! I'm still trying... Two daft approximations that I have done: $$F_{m} = \frac{F_{o} + F_{f}}{2} = \frac{F \cdot \cos (37º) + F\cdot \cos (53º)}{2} = F\cdot \frac{7}{10}$$; therefore, $$ W = F_{m} \cdot d = \frac{7}{10} \cdot F \cdot d$$, which is somewhat close to 5Fd/7... A better...

https://www.khanacademy.org/math/multivariable-calculus/integrating-multivariable-functions/line-integrals-vectors/v/using-a-line-integral-to-find-the-work-done-by-a-vector-field-example I was watching this video about calculating work using integration. But in this example, F is written in...

$$ W = \int \vec{F} \cdot d\vec{d} = \vec{F} \int d\vec{d} = \vec{F} \cdot \vec{d} = F \cdot d \cdot \cos \theta$$, right? But $$\theta$$ ranges from 37º to 53º, which the definite integral would give sin(53º) - sin (37º) = 4/5 - 3/5 = 1/5 (because the $$\int \cos(x) dx = \sin(x)$$); therefore W...

Homework Statement A force F acts over the block as show in the figure. F have a constant magnitude, but has a variable direction. The direction of F, at any given instant, always is directed to the point P. Evaluate the work done by the force F through the displacement d between the points R...

Thank you very much - for the welcome and for the solution. :) After your comment, I noticed that in my figure the tension and the weight have similar lenghts. I redone it with more precise lenghts (T = 2W) and, indeed, it's impossible to the resultant vector to be horizontal in terms of...

Homework Statement A block of mass ## m ##, attached to a rope, is dropped at the point A. When the block reaches the point B, the tension is ## T = 2 \cdot m \cdot g ## and the rope broke at that same point. If the length of the rope is ## L = 6cm ##, evaluate the height ## h ## where the...