Problem 6/12  Page 430, Engineering Mechanics - Dynamics, Meriam and Kraige, 4th Edition

Given: A cabinet of mass M resting on a cart of mass MC.  The cabinet rests on legs separated by a distance L.  The cabinet mass center is midway between the two legs and a distance of H above the legs.  The cart is accelerated along a horizontal surface by a force P.  The cart is mounted on frictionless wheels.
 
Find:  The value of the force P that would cause the cabinet to tip.  Determine the static coefficient of friction necessary to ensure that tipping occurs without slipping.
 
Parameter Values:
M = 50 kg = Mass of cabinet.
MC = 10 kg = Mass of cart.
L = 0.8 m = Base of cabinet
H = 0.6 m = Height of cabinet mass center.

1. Mechanical Systems: Two mechanical systems are required.  The first is the cabinet.  The second is the cart.  Both are considered during the time the force P is applied.

2. Free Body Diagrams: Motion occurs horizontally to the right.  Thus we will label this the t direction.  The n direction will be vertically upward.  The cabinet free body diagram shows the weight of the cabinet acting vertically downward.  The cabinet is about to tip around its left leg so there are no forces at the right leg.  At the left leg we have a normal force acting vertically upward and a frictional force acting to the right.  These leg forces are exerted by the cart on the cabinet so the law of action and reaction must be applied when constructing the free body diagram of the cart.  The cart free body diagram showns the left leg normal force acting vertically downward and the left leg friction force acting to the left.  The cart free body diagram also shows its weight acting vertically downward and the applied force P acting horizontally to the right.  Finally the cart free body diagram showns three normal forces acting vertically upward on the three cart wheels (no frictional forces are shown because the wheels are presumed to be light and to turn freely).

3. Equations:
a) Kinetics:
Cabinet:
S F = F t + ( N - M g ) n
S M_G = ( -H n - L/2 t )  X ( F t + N n )
          = ( H F - L/2 N ) k
Cart:
S F = ( P - F ) t + ( N1 + N2 + N3 - MC g - N ) n

b) Kinematics:
Cabinet
m a_G = M a_t t
(Cabinet and cart are moving in a straight line and hence have no normal acceleration).
I_G a = 0
(Cabinet is not yet tipping, so it is not rotating).
Cart
m a_G = MC a_t t
(Cabinet and cart are moving together so both must have the same tangent acceleration).

4. Solve:
F = M a_t
N - M g = 0
H F - L/2 N = 0
P - F = MC a_t
From the second equation:
N = M g
From the third equation:
F = L/(2 H) N = L/(2 H) M g
From the first equation:
a_t = F / M = L/(2 H) g
From the fourth equation:
P = F + MC a_t = (M + MC) L/(2 H)  g
P = 392 N
m_s = F/N = L/(2 H) = 0.667