# Solve Sliding Dresser Problem - Find F, Np, Nq

• whitetiger
In summary, the problem involves finding the magnitude of the force F, the normal forces Np and Nq, and the torque for a dresser sliding on the floor with a coefficient of kinetic friction μk. By using the equations of summation of forces and torques, and substituting for known values, the equations can be solved to find the values of F, Np, and Nq. The final steps involve substituting these values back into the equations to solve for the remaining variables.
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)

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

OlderDan said:
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)

whitetiger said:
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.

OlderDan said:
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...

whitetiger said:
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.

OlderDan said:
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

whitetiger said:
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 . . . .

OlderDan said:
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

Last edited:
whitetiger said:
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.

## 1. What is the "Sliding Dresser Problem"?

The "Sliding Dresser Problem" is a physics problem that involves finding the forces acting on a dresser as it slides down an inclined plane. It is used to understand the principles of mechanics and the relationships between forces, mass, and motion.

## 2. What do F, Np, and Nq stand for in this problem?

F represents the force of gravity acting on the dresser, Np represents the normal force exerted by the inclined plane on the dresser, and Nq represents the normal force exerted by the floor on the dresser. These forces are all important in determining the motion of the dresser down the inclined plane.

## 3. How do you solve the "Sliding Dresser Problem"?

To solve the "Sliding Dresser Problem", you need to use the equations of motion and the principles of mechanics to determine the forces acting on the dresser. This involves setting up free body diagrams, applying Newton's laws of motion, and solving for the unknown variables (F, Np, and Nq).

## 4. What factors affect the forces in the "Sliding Dresser Problem"?

The forces in the "Sliding Dresser Problem" are affected by the mass of the dresser, the angle of the inclined plane, and the coefficient of friction between the dresser and the plane. These factors can change the magnitudes and directions of the forces, ultimately affecting the motion of the dresser.

## 5. Why is the "Sliding Dresser Problem" important in science?

The "Sliding Dresser Problem" is important in science because it helps us understand the fundamental principles of mechanics and how forces interact with objects. It also has real-world applications in fields such as engineering and physics, where understanding and manipulating forces is crucial. By solving this problem, we can gain a better understanding of the physical world around us.

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