The problem involves determining the coefficient of kinetic friction for a wooden crate being pushed across a floor with a constant horizontal force. The crate has a mass of 12.0 kg, and a force of 30.0 N is applied to maintain its motion at constant velocity.
Several participants have attempted calculations and shared their reasoning, but there is no clear consensus on the correct approach. Some have suggested that the frictional force must equal the applied force for the crate to move at constant velocity, while others are still trying to reconcile their calculations with the problem's requirements.
Participants are working under the assumption that the crate is moving at constant velocity, which implies that the net force acting on it is zero. There is also discussion about the normal force and its relationship to the gravitational force acting on the crate.
You need to identify all the forces acting on the crate, and then apply Newton's laws in both the x and y direction.ilkjester said:Homework Statement
If you use a horizontal force of 30.0 N to slide a 12.0-kg wooden crate across a floor at a constant velocity, what is the coefficient of kinetic friction between the crate and the floor.
Homework Equations
Fkf=μFn
The Attempt at a Solution
Im not sure how to find the μ
You got 117.6 for which force? Please show your work.ilkjester said:Ok well i got 117.6 when i did the problem not sure if its right or not.
aww man and i thought i had something. Well I guess ill have to keep trying something I don't know why I am not getting this problem the rest of them were easy i think once i get it I am going to be like duhh.pooface said:no 117.72 N is the Normal force on the crate.
ilkjester said:Homework Statement
If you use a horizontal force of 30.0 N to slide a 12.0-kg wooden crate across a floor at a constant velocity, what is the coefficient of kinetic friction between the crate and the floor.
Homework Equations
Fkf=μFn
The Attempt at a Solution
Im not sure how to find the μ
ilkjester said:Kinetic friction force Ff kinetic =μkFn
so would it be 6=μk(30N)Sourabh N said:You can find Fn by balancing forces in vertical direction. Now frictional force is balanced by external force (30 N). Thus you can find \mu_{k}
Cheers
ilkjester said:ok i think I've got this now. But its assuming that the floor is wooden.
Kinetic friction force Ff kinetic =?kFn
Now ?k wood on wood equals 0.20
so Ff kinetic=0.20(30)
Ff kinetic=6
does this sound right.
ilkjester said:so would it be 6=?k(30N)
then divide by 30 on both sides so 6/30=.2?k
D H said:You are supposed to compute u_k, not look it up. And no, this is not correct. The crate will accelerate at 24N/12kg=2 m/s2 if a person applies a 30N horizontal force in one direction and friction only applies a 6N force in the opposite direction.
You are making this too hard. The crate is moving at a constant velocity. This means that the net horizontal force acting on the crate is zero. The person is applying a 30N force in one direction. To make the net horizontal force zero, friction must therefore be applying a 30N force in the opposite direction. The coefficient of kinetic friction is just the ratio of the frictional force (which you now know) to the normal force (which you calculated in post #4).
so was i right in my one post thenilkjester said:so would it be 6=μk(30N)
then divide by 30 on both sides so 6/30=.2μk
How did you get that out of what I wrote?? The frictional force is 30N.ilkjester said:so was i right in my one post then