Sliding dresser

[whitetiger]

http://img244.imageshack.us/img244/1565/mfsto2ie5.jpg

Hi all,

Would some one please take a look at this problem for me? I am trying to work out this problem, but I have some trouble...

The coefficient of kinetic friction between the dresses and the floor is μk. The ground exerts upward normal forces of magnitudes Np and Nq at the ends of the dresser.

a) If the dresser is sliding with constant velocity, find the magnitude of the force F.
b) Find the magnitude of the normal force Np
c) the magnitude of the normal force Nq.

I have tried to work our part of the problem, but I am not sure.

summation Fy = nq - w = 0 so Nq = w
summation Fx = F - f = 0 so F =f
F = μkw so F = μk(mg)
not sure if this is correct, the magnitude of F is
F = μk(mg)

Please help,

[OlderDan]

You left out one of the forces in your y equation. To find the N_p and N_q you need to use torque.

[whitetiger]

You left out one of the forces in your y equation. To find the N_p and N_q you need to use torque.

Thank for replying to my question. I am not sure how to go about it. Can you help? I can further found that the torque for point at Q is

summation torque Q = rwsin(theta)

[OlderDan]

Thank for replying to my question. I am not sure how to go about it. Can you help? I can further found that the torque for point at Q is

summation torque Q = rwsin(theta)
There is no acceleration in the problem, so the sum of the forces must be zero. There are three vertical forces and three horizontal forces. You combined two of the horizontal forces in your first post, and you can get away with that in this problem, but it would be good to keep them separate.

There is no rotation in the problem, so the sum of the torques about any point is zero. One of the points of contact with the floor would be a good place about which to calculate the torques. Point Q is a good choice.

[whitetiger]

There is no acceleration in the problem, so the sum of the forces must be zero. There are three vertical forces and three horizontal forces. You combined two of the horizontal forces in your first post, and you can get away with that in this problem, but it would be good to keep them separate.

There is no rotation in the problem, so the sum of the torques about any point is zero. One of the points of contact with the floor would be a good place about which to calculate the torques. Point Q is a good choice.

I am total stuck,can someone help please....

[OlderDan]

I am total stuck,can someone help please....
List the forces you think are acting vertically. List the forces you think are acting horizontally. List the torques you think are acting about point Q, and we'll take it from there.

[whitetiger]

List the forces you think are acting vertically. List the forces you think are acting horizontally. List the torques you think are acting about point Q, and we'll take it from there.

sorry for the delay,

here is what I have so far

Fy = mg = Np +Nq
Fx =f= μk(Np +Nq)
Torque is Fh +NpL - mgL/2 = 0

[OlderDan]

sorry for the delay,

here is what I have so far

Fy = 0 » mg = Np + Nq
Fx = 0 » F = f = μk(Np + Nq)
Torque is Fh +NpL - mgL/2 = 0
I know what you mean, but technically you have misused the equal sign in your first two equations so I changed your statement a bit. You have all the needed information in this set of equations except that the magnitude of the applied force equals the magnitude of the friction force (you had that in your original post), so I added that to the equation.

You now have three equations for the three unkowns: F, Nq and Np. All you need to do is solve the system of equations. Can you so that? The first two equations will give you F easily. You can then sustitude F into the last equation to solve for Np. Then . . . .

[whitetiger]

I know what you mean, but technically you have misused the equal sign in your first two equations so I changed your statement a bit. You have all the needed information in this set of equations except that the magnitude of the applied force equals the magnitude of the friction force (you had that in your original post), so I added that to the equation.

You now have three equations for the three unkowns: F, Nq and Np. All you need to do is solve the system of equations. Can you so that? The first two equations will give you F easily. You can then sustitude F into the last equation to solve for Np. Then . . . .


This is what I have for the problem:
Fy = 0 » mg = Np + Nq =======> mg = Np + Nq
Fx = 0 » F = f = μk(Np + Nq) =====> F = f = μk(Np + Nq) , since mg = Np + Nq, we can replace Fx equation with mg. This will give us F=f=μk(mg)

Replacing the F equation back into the torque equation Fh +NpL - mgL/2 = 0 and we get Np = (mg(L/2)-μk(mg)h)/L

After getting Np value, we substitute that back into mg = Np + Nq and solve for Nq

Please correct me if I am wrong.

Thank you very much for your help. I am now able to calculate for all the required variables.

Have a good day,
WT

[OlderDan]

This is what I have for the problem:
Fy = 0 » mg = Np + Nq =======> mg = Np + Nq
Fx = 0 » F = f = μk(Np + Nq) =====> F = f = μk(Np + Nq) , since mg = Np + Nq, we can replace Fx equation with mg. This will give us F=f=μk(mg)

Replacing the F equation back into the torque equation Fh +NpL - mgL/2 = 0 and we get Np = (mg(L/2)-μk(mg)h)/L

After getting Np value, we substitute that back into mg = Np + Nq and solve for Nq

Please correct me if I am wrong.

Thank you very much for your help. I am now able to calculate for all the required variables.

Have a good day,
WT
Looks good. I think you got it.:smile: