2 blocks on top of each other, applied force, friction, and tension

AI Thread Summary
A 4.32 kg block is placed on a 9.16 kg block, with a horizontal force of 66.6 N applied to the lower block while the upper block is tied to a wall. The coefficient of kinetic friction is 0.169, affecting both blocks. The initial calculations for tension in the string were incorrect, as the applied force on the bottom block does not directly influence the top block. The correct approach identifies that the tension equals the frictional force acting on the upper block, resulting in a tension of 7.155 N. This clarification resolves the misunderstanding regarding the forces acting on the upper block.
TFOrange
Messages
3
Reaction score
0

Homework Statement


A 4.32 kg block is placed on top of a 9.16 kg
block. A horizontal force of F = 66.6 N is
applied to the 9.16 kg block, and the 4.32 kg
block is tied to the wall. The coefficient of
kinetic friction between all moving surfaces
is 0.169. There is friction both between the
masses and between the 9.16 kg block and the
ground.
The acceleration of gravity is 9.8 m/s^2.
Determine Tension of string in N

(see image)

Homework Equations


ƩF = ma
fs = μk*FN


The Attempt at a Solution


(see attachment for free body diagrams)
ƩF = ma
m = 4.32 kg
atop block = 0
ƩF = Fapplied - fk - T
0 = (44.274) - (7.155) - T
T = 44.274 - 7.155
T = 37.119

Apparently I did something wrong somewhere, but I'm not sure where.
 

Attachments

  • PF.png
    PF.png
    16 KB · Views: 1,888
  • PF2.png
    PF2.png
    8.8 KB · Views: 1,309
Last edited:
Physics news on Phys.org
What forces act on the upper block? (I don't understand your FBDs.)
 
Well maybe it isn't supposed to be there, but I was thinking that since the bottom block has a force in that direction that is not completely canceled out by friction, it would cause the top block to have a force as well, which is equal to Fapplied (to the bottom block) - fk
 
TFOrange said:
Well maybe it isn't supposed to be there, but I was thinking that since the bottom block has a force in that direction that is not completely canceled out by friction, it would cause the top block to have a force as well, which is equal to Fapplied (to the bottom block) - fk
That applied force acts on the bottom block, not on the top block. So it doesn't belong in your analysis of the top block. The only influence that the bottom block exerts on the top block is through the normal force and friction.
 
Oh! so T = f
and so it's T = 7.155
Thanks!
 
TFOrange said:
Oh! so T = f
and so it's T = 7.155
Yep!
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top