Block A in Figure 5.62 http://img222.imageshack.us/img222/3261/figure562sg6.jpg weighs 1.40 N, and block B weighs 4.20 N. The coefficient of kinetic friction between all surfaces is 0.30. Find the magnitude of the horizontal force F necessary to drag block B to the left at constant speed if A and B are connected by a light, flexible cord passing around a fixed, frictionless pulley.
2006-08-26 23:28:43 · 3 answers · asked by just_askin' 1 in Science & Mathematics ➔ Physics
The combined weight of A and B is 1.4+2.2=5.6 N which is also the reaction of the surface on which A moves and the corresponding frictional force opposing the applied force is 5.6*0.3=1.68 N.
Reaction of B on A is 1.4 N and the corresponding frictional force is 1.4*0.3=0.42 N.
The tension in the cord is therefore 0.42 N and B is pulled by this tension in the opposite direction to the applied force.
Hence the force required to move block B with a constant velocity is 1.68+0.42=2.1 N
2006-08-27 07:10:39 · answer #1 · answered by rabi k 2 · 2⤊ 3⤋
I agree with John's reasoning but not his calculation:
(4.2 + 1.4) * .3 = 1.68 N, I think.
How could it take 18 Newtons to pull two tiny weights such as 4.2 and 1.4 Newtons? How could the force needed be greater than the weight of the objects?
2006-08-27 00:46:52 · answer #2 · answered by ? 6 · 0⤊ 3⤋
Just add the force to drag each block on its own.
18.7N
2006-08-26 23:39:18 · answer #3 · answered by john a 2 · 0⤊ 2⤋