The problem involves determining the magnitude of a pushing force on a block moving at constant velocity, with a given coefficient of kinetic friction. The context includes concepts of forces, friction, and equilibrium conditions in physics.
Participants are actively engaging with the problem, clarifying concepts related to equilibrium and the relationships between forces. Some guidance has been provided regarding the equations to use, but there is no explicit consensus on the final expression for the pushing force.
There is a noted confusion due to the absence of mass information, which is essential for calculating the forces involved. The discussion also reflects on the implications of constant velocity and the resulting net forces being zero.
jr4life said:no they didnt give the mass of the block, that is why I am so confused my teacher said something about sigma Fx=0 and sigma Fy=0 but i don't know what to do with those numbers, all i know is velocity is constant which makes acceleration 0 which makes force of x and y 0, but after that I am lost
jr4life said:In the x direction is the pushing force and force of friction and in the y direction is force of gravity on a box whos mass is not given and the normal force, i think the lack of given numbers is confuing me
jr4life said:in x direction it would be (mass)(0) so resultant force would be 0, because acceleration would be 0 due to constant velocity and in y direction (mass)(9.8) and then i don't really know what you could do with that...
learningphysics said:Write the equations using Fpushing, friction, mg, Fnormal... how do they add or subtract?
jr4life said:mg-Fnormal=0
this is where i get confused with x because they are not in equilibrium so they wouldn't equal 0
Fpushing-friction=?? idk
learningphysics said:It equals 0, because acceleration is 0 (constant velocity). Now, you also know that friction = [tex]\mu * F_{normal}[/tex]
Solve for Fpushing using your two equations, in terms of mass, [tex]\mu[/tex] and g.
jr4life said:ok, i understand why it equals zero now but when you use that equation would it be
friction=(0.5)(9.8m) and which two equations are you referring to?
learningphysics said:Yes, that's right...
The two equations I meant were:
Fpushing - friction = 0
Fnormal - mg = 0
you actually used the second equation to get friction = 0.5*9.8*m
So what does Fpushing come out to...
jr4life said:Fpushing-(0.5)(9.8m)=0
Fpushing-4.9m=0
Fpushing=4.9m??
FedEx said:Seems good to me.