The problem involves calculating the force required to move a 50kg crate resting on the floor, given a coefficient of static friction of 0.5. Participants are exploring the relationship between frictional force and normal force in the context of static friction.
The discussion is progressing with participants attempting to calculate the normal force and frictional force. Some guidance has been provided regarding the use of the equation and the relationship between mass, weight, and friction. Multiple interpretations of the problem are being explored, particularly regarding the calculations involved.
Participants are working under the constraints of a homework assignment, which may limit the information they can reference or the methods they can use. There is ongoing questioning about the correct application of the equations involved.
rock.freak667 said:Do you know the equation that F=uR, where F is the frictional force and R is the normal reaction?
rock.freak667 said:Firstly, you'll need to find the normal reaction,R. In this case it is simply the crate's weight. When you have that, put that into F=uR where u is the coefficient of friction. Then compare the value of friction that you get to the answers given.
vballkatie22 said:Ok. That sounds pretty easy. So for that it would look like this..
F= uR
F= (.5)(50)
I guess I multiply right? And that gives me 25N but that isn't an answer.
Did I do something wrong?
rock.freak667 said:50kg is the crate's mass. To find its weight you need to multiply the mass and the acceleration due to gravity
vballkatie22 said:Ok. So to find the weight I take 50kg and times that by acceleration which is 9.8 (gravity)
To get 490.
Correct?
rock.freak667 said:yes that is the weight of the crate and hence its normal reaction. Now what is the frictional force (this also is the frictional force which causes the crate to be in limiting equilibrium)
vballkatie22 said:Would it be .5? I don't really know.