Hand Pushes Block: Horizontal Forces Explained

  • Thread starter Thread starter mujadeo
  • Start date Start date
  • Tags Tags
    Block hand
AI Thread Summary
The discussion focuses on understanding the forces acting on a block after a hand has pushed it and subsequently let go. It clarifies that while kinetic friction acts opposite to the block's motion, there is no additional force in the direction of motion once the hand is no longer in contact. The block continues to move due to its initial velocity, as forces are not required to maintain motion but rather to change it. The conversation emphasizes that a free body diagram (FBD) illustrates forces but does not indicate the direction of motion, which is determined by initial conditions and kinematic equations. Ultimately, the net force can be zero while the block is still in motion, highlighting the distinction between force and velocity.
mujadeo
Messages
103
Reaction score
0

Homework Statement


Not an actual HW problem, but related to a lot of probs I'm doing.
Lets say a hand pushes a block across a surface (with friction).
You take pic of 2 timepoints
1) when hand pushes block
2) After hand has let go and block is slowing due to friction

Now when I draw the FBD for situation #2, I am confused about the horizontal forces. (i know that vertical forces cancel)

I know that 1 vector (kinetic friction) will point opposite the motion.



Homework Equations


BUT: Is there a forces that goes in the direction of motion?
If there isn't, then how is the block moving??
If there is, then what force could that be, seeing as the hand is no longer in contact with the block??


Thanks in advance for any explanation!

The Attempt at a Solution


 
Physics news on Phys.org
mujadeo said:

Homework Statement


Not an actual HW problem, but related to a lot of probs I'm doing.
Lets say a hand pushes a block across a surface (with friction).
You take pic of 2 timepoints
1) when hand pushes block
2) After hand has let go and block is slowing due to friction

Now when I draw the FBD for situation #2, I am confused about the horizontal forces. (i know that vertical forces cancel)

I know that 1 vector (kinetic friction) will point opposite the motion.
yes, correct.



Homework Equations


BUT: Is there a forces that goes in the direction of motion?
What does your FBD show you?
If there isn't, then how is the block moving??
The hand sure helped. What would happen to the object after the hand was released if there were no friction? Why?
If there is, then what force could that be, seeing as the hand is no longer in contact with the block??
Right, good point.
 
My problem is in drawing the correct FBD, so at moment I am not relying on FBD for info
But this is how i think it should look: (W and N are sposed to be equal lengths)

http://www.imagination3.com/LaunchP...l=mujadeo@yahoo.com&to_name=&_lscid=184982417

Yes I know hand got it moving, but if FBD is sposed to show forces on it, then there should be no force in pos x direction (right), because the hand is no longer in contact with block, right?
But if no force in in pos x, then according to diagram, Fnet is to the left, and object is moving to the left (according to the diagram)?

If there was no friction, then block would move at constant velocity in pos x direction. (and accel would be zero)

tanks for your help
 
Last edited:
mujadeo said:
My problem is in drawing the correct FBD, so at moment I am not relying on FBD for info
But this is how i think it should look: (W and N are sposed to be equal lengths)

http://www.imagination3.com/LaunchP...l=mujadeo@yahoo.com&to_name=&_lscid=184982417

Yes I know hand got it moving, but if FBD is sposed to show forces on it, then there should be no force in pos x direction (right), because the hand is no longer in contact with block, right?
But if no force in in pos x, then according to diagram, Fnet is to the left, and object is moving to the left (according to the diagram)?

If there was no friction, then block would move at constant velocity in pos x direction. (and accel would be zero)

tanks for your help
Yes, your FBD is OK. You have correctly noted that there is only one horizontal force acting, the friction force, which is the net force, acting left. But per Newton 2, if the net force is left, then the acceleration is left, not necessarily the motion, which in this case is still to the right until the block stops. As you also have correctly noted, you don't need forces to keep a body moving. Forces retard or accelerate the motion; they do not keep it in motion, as noted in Newton 1.
 
SO am i corerect in saying that A FBD tells us nothing about which direction an object is moving (the velocity vector) , only its direction (and mag) of acceleration.
If that's true then Fnet says nothing about velocity, and a system with Fnet=0 could be stationary or going any speed or direction? ( velocity is completely independent of Fnet at all right?)
 
mujadeo said:
SO am i corerect in saying that A FBD tells us nothing about which direction an object is moving (the velocity vector) , only its direction (and mag) of acceleration.
If that's true then Fnet says nothing about velocity, and a system with Fnet=0 could be stationary or going any speed or direction? ( velocity is completely independent of Fnet at all right?)
That is correct. You get velocity from the kinematics or energy equations, or given data, etc. A system with F_net = 0 is either at rest or moving at constant speed in a certain direction, and will remain at rest, or moving in a straight line at a constant speed in that direction, as the case may be, unless acted on by a net unbalanced force (Newton 1).
 
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