Recent content by alexi_b

im still lost, I don't know what to do :(, any more hints?

yeah but i used those measurements to calculate the angle at which i was pulling at, no?

yes that is what i meant

the angle came from using the length and height given in the problem

Oh right! Sorry i calculated too quickly Fnet = Fn + Tsin(theta) - Fg 0 = Fn + Tsin(theta) - Fg Fg - Tsintheta = Fn 23.1 N - Tsin17.3 degrees = Fn 23.1 - 0.297T = Fn

right so i got 23.1 N, but I still have no idea how to continue this problem

It would be Fnet = Tcos(theta) - kinetic friction (as they act in opposite directions) then for calculating frictional for I would use F(kinetic)=u*N right?

kinetic friction and tension

Homework Statement A block of mass 2.42kg is accelerated across a rough surface by a rope passing over a pulley, as shown in the figure below. The tension in the rope is 13.9N, and the pulley is 12.3cm above the top of the block. The coefficient of kinetic friction is 0.395. (a: 2.28m/s^2)...

so if i calculate the normal force of the top block to be 22.0 N, since there are two normal forces acting on the bottom block, the normal force of the top block will act as one of the normal forces of the bottom block. Should it be 22.0 N or -22.0 N?

So if i calculate the force of the normal from the top block, according to the diagram it is the same magnitude for the bottom block? Or do i reverse its signs?

yes i am so sorry ;(! that's why I was getting confused because I thought that the tension in the rope = static friction of the top block. I am back to square one now, please help!

yeah it is. But what i wish i could show you is the diagram. I should probably mention that the top block has a rope attached to it and this is where that question i asked you about calculating static friction on its own came from. Knowing this, would it affect the outcome?

So i get: Fs = (coefficient of static friction) x (normal force of top block) Fs = 13.3 13.3 = mb(a) 13.3= 2.25kg (a) a = 5.91 m/s^2 d = Vit + 1/2at^2 (Vit vanishes because there's no initial velocity) d = 1/2at^2 3.44 = 1/2 (5.91)t^2 3.44/2.955 = t^2 //sqrt both sides 1.08 s = t this...

Homework Statement A block of mass 2.42kg is accelerated across a rough surface by a rope passing over a pulley, as shown in the figure below. The tension in the rope is 13.9N, and the pulley is 12.3cm above the top of the block. The coefficient of kinetic friction is 0.395. A: 2.28 m/s^2...