Question

if a block is being pulled by a rope angled upwards, does the normal force include the vertical tension of the rope?

Answer 1

suppose the mass was m, the force of rope F, the angle above the horizontal of the force is T and gravity is g... how would i calculate the normal force?

Answer 2

its been a while since ive done any physics and im being asked on the basics of AP physics c again... the free response are killing me

Answer 3

If we're talking about a block on a level surface with the rope angled up, part of the force applied through the rope (the vertical component) counteracts part of the gravitational force on the block and so the normal force (the force applied by the surface onto the block) is less than the force of gravity on the block.

Does that make sense? I can show it mathematically if need be...

Answer 4

yes that makes sense, so the normal will be n-the vertical tension so n-FsinT ?

Answer 5

jeez, and i was thinking that i would add the normal and vertical tensional forces -.-

Answer 6

oh i probably should say mg-FsinT instead of using n

Answer 7

Yup :)

But I think it would be clearer if you wrote N = G - Tsinθ, where N is the normal force, G is the gravitational force, and Tsinθ is the vertical component of the tension.

Answer 8

Slash you just wrote that so yeah...

Answer 9

is capital g (G) considered the force of gravity rather than the acceleration or coefficient?

Answer 10

I was just using that here...in general you'd be most safe using mg or Fg (g is subscript)

Answer 11

Yeah you're right you wouldn't want to confuse that with the gravitational constant lol

Answer 12

also, the acceleration

Answer 13

if the crate had a coefficient of kinetic friction, u, then wouldnt the acceleration be a= (FcosT-Fu)/m ?

Answer 14

where the F in FcosT is the tensional and Fu is the frictional forces

Answer 15

a = (Tcosθ - f)/m

If that's what you meant then yes

Answer 16

yes, thats what i meant. so it is divided by the mass! thanks!

Answer 17

Np