Recent content by Hemmelig

Hey Working on a rather large project and it's the first time I've had to look into the world of pressure sensors. basically, we're looking into measuring flow levels in small rivers, either in areas where we build a profile to gather up the water and in natural areas where there's no need...

But he's pushing down at the box with an angle of 30 , won't the frictional force be horizontal, as in f x cos (180) then ?

My educated guess would be that I'm wrong Homework Statement A factory worker pushes a 30kg crate a distance of 4,5m along a floor at constant velocity, the coefficient of kinetic friction between the crate and floor is 0.25 The angle he pushes the crate with is 30 degrees below the...

Great :) Again, thank you so much

Ah, ofcourse From equation a i have that T=12,753 - 1,3a From equation b i have that T=2,25a + 9,93 I solve for T and get a=2,823/3,55 = 0,7952 I then put it into equation a and get t=11,71 I can then use V^2 = V0^2 + 2a (x-x0) to find the speed after 0.03m (is that the best equation to use...

Hmm, so what do i do next ? I don't know T and i don't know a

Hmm \sum F_B = mg - T = m_B\cdot a ?

Well My guess basically brings me down the same route as before Can i say: Sum of all forces F= sum of Force(a) + sum of force(b) = 2.823N Then i can just use F=ma to find a ? by dividing F with the total mass of the two blocks But that's basically what i did earlier on, which was wrong

Hmm, i don't really know what to do next It would be more of a guess than actual knowledge :/

The tension in the rope connecting the blocks is \mu * m(a)g So the sum of all forces working on block B is it's weight trying to pull it down, minus the tension of the rope ?

Hmm Well, it's weight is trying to pull block B down, while Block A and the friction is basically holding back.

Hmm, the two forces acting on B is it's mass and g. They are acting downwards, which is in the same direction of the motion of B Right ?

Hmm, I'm not sure what the second block will be \sum F_B= N + (-mg) ?

I see I'm on my way out, so i'll try when i get back in Just one more thing, i tried doing this and i got the right acceleration, but i don't really know if I'm allowed to do it I know the frictional force fk that is trying to hold the blocks back I know the force that block B is trying to...

The sum of forces working on Block A is: the normal force n, which is the same as mg fk = the frictional force that tries to hold it back then you have the force from block B which is trying to pull block A