Sam is trying to move a dresser of mass m...?http://session.masteringphysics.com/problemAsset/1010956/19/MFS_to_2.jpg

Sam is trying to move a dresser of mass m and dimensions of length L and height H by pushing it with a horizontal force F_vec applied at a height h above the floor. The coefficient of kinetic friction between the dresser and the floor is mu_k and g is the magnitude of the acceleration due to gravity. The ground exerts upward normal forces of magnitudes N_P and N_Q at the two ends of the dresser. Note that this problem is two dimensional.

A) If the dresser is sliding with constant velocity, find F, the magnitude of the force that Sam applies.Express the force in terms of m, g, and mu_k.

B) Find the magnitude of the normal force N_P. Assume that the legs are separated by a distance L, as shown in the figure.Express this normal force in terms of m, g, L, h, and mu_k.

C) Find the magnitude of the normal force N_Q. Assume that the legs are separated by a distance L, as shown in the figure.Answer in terms of m, g, mu_k, h, and L.

D) Find h_max, the maximum height at which Sam can push the dresser without causing it to topple over.Express your answer for the maximum height in terms L and mu_k.

Answer

..................L

.........▓▓▓▓▓▓▓▓▓

.........▓▓▓▓▓▓▓▓▓..▬▬►

.........▓▓▓▓▓▓▓▓▓..(unbalanced force)

.........▓▓▓▓▓▓▓▓▓

.F==>▓▓▓▓▓▓▓▓▓.H

...↑ ...▓▓▓▓▓▓▓▓▓

...h....▓▓▓▓▓▓▓▓▓

.........▓▓▓▓▓▓▓▓▓

.. ↓__.▓▓←Pf.....▓▓←Qf

.........↑ ................↑

.........Np..............Nq

Pf =µĸ Np <----- (frictional force at P}Qf =µĸ Nq <-----(frictional force at Q)

A) Unbalanced force = F - Pf - Qf

........F must be more than (Pf + Qf )

F - Pf - Qf = maF = ma + Pf + Qf...=ma + µĸ Np + µĸ Nq

B) Solving for the value of Np

+↑(-)↓ ∑Mq = 0

0 = Fh- mg(L/2) + Np(L)Np(L) = - Fh + mg(L/2)Np = - F(h/L) + mg(1 /2)

Np = ½ mg - F(h/L) ◄

C) Solving for Nq

+↑(-)↓ ∑Mp = 0

0 = Fh + mg(L/2) - Nq(L)Nq(L) = Fh + mg(L/2)Nq = Fh(1/L) + mg(1/2)

Nq = F(h/L) + ½ mg ◄

D) For the condition when the dresser is about to topple over, the dresser will be lifted at point P, hence Np = 0

The maximum h will be obtained by letting Np = 0 and the dresser will pivot at point Q

+↑(-)↓ ∑Mq = 0

0 = Fh - mg(L/2)Fh = mg(L/2)

h = ½ mg (L / F) ◄