Note: We are displaying rounded intermediate values for practical purposes. However, the calculations are made using the unrounded values.

The free-body diagram for each system is shown below.

N

is the normal force exerted by the surface on the system,

F_C

is the contact force,

f

is the friction force and

W

refers to the weight. Note that for the system containing both boxes, the contact force

F_C

is not external to the system. And when you consider just box B, the force

F_P

does not act on it.

(a) To find the acceleration of the two boxes, consider both boxes as the system. Apply Newton's Second Law

F_net = ma

to this system for motion in the horizontal and vertical directions.

vertical direction:
Recall that f = μN and therefore
horizontal direction:
(b) To find the contact force F_C exerted on box B by box A, consider box B as the system. Now the friction force depends on the normal force exerted by the ground on box B alone. Apply Newton's Second Law to box B.
(c) If Alex now pushed on box B rather than on box A, you would have to switch the two masses in the solutions above. Therefore the contact force would now be larger since mass of A is larger. We can check this out by performing the calculations.

(a) If we consider both boxes to be the system of interest, the tension is an internal force and the net work done by the tension on the system is zero. As a result we may use conservation of energy to determine the distance. This allows us to write

ΔKE = −ΔPE.

Where Inserting expressions for the initial kinetic energy of each object, obtain: where we have set

v_1i = v_2i = v

since they are moving with the same speed. The expression for the change in potential energy is Inserting expressions for ΔKE and ΔPE into

ΔKE = −ΔPE,

obtain: Solving for d, obtain:
(b) Since the tension acting on box 2 is in the direction of motion for box 2, the tension does positive work on box 2.

(c) The work done by nonconservative forces (the tension) acting on box 1 is the change in total mechanical energy of box 1. This allows us to write the work done on box 1 by the tension as: We expect a negative value since the force (tension in the rope) acts in the direction opposite that of the motion.

Note: We are displaying rounded intermediate values for practical purposes. However, the calculations are made using the unrounded values.

(a) The figure below shows all forces acting on the ladder when the person is halfway up the ladder. As a result of the first condition of equilibrium, we know that the sum of all forces acting upward on the ladder is equal to the sum of all forces acting downward. This allows us to write As a result of the first condition of equilibrium, we know that the sum of all forces acting on the ladder to the left is equal to the sum of all forces acting to the right. This allows us to write We cannot determine f₁ by the expression

f₁ = μN₁,

because we do not know the coefficient of friction between the ground and ladder. As a result, let's see if we can use the second condition of equilibrium, the basic definition of torque, and the base of the ladder as the point to take the torques about in order to determine the normal force N₂. From the second condition of equilibrium we can state that the sum of the clockwise torques about a point at the base of the ladder is equal to the sum of the counterclockwise torques, or

τ_cw = τ_ccw.

The weight of the ladder and the person provide clockwise torques about a point at the base of the ladder and the normal force from the wall provides the counterclockwise torque. Using the basic definition of torque in the form

τ = Fr_⊥

and inserting details we may write where r_⊥ for the force

(W_L + mg)

is

(L/2)cos θ

and r_⊥ for the force N₂ is d. Solving this expression for N₂, we obtain Before we can obtain a numerical value for N₂, we need to obtain a value for the angle θ. From the figure we see that Inserting this and other values into the expression for N₂, obtain Inserting values we have Now that we know N₂, we may write a horizontal first condition of equilibrium statement that will allow us to determine f₁.
(b) The figure below shows all forces acting on the ladder when the person is three fourths the way up the ladder. Notice that the upward and downward forces acting on the ladder are the same as for part (a). Granted the weight of the person is now acting on the ladder at a different place, but the upward and downward forces are the same, hence the vertical statement of the first condition of equilibrium is the same, and we still have Looking at the figure, the only thing that has changed is the location of the weight of the person and hence r_⊥ for the weight of the person. Summing torques about the base of the ladder, obtain Again, from the horizontal statement of the first condition of equilibrium we may write

Since the object is in vertical equilibrium, we know that the total buoyant force acting on the object is equal to the force of gravity acting on the object. This allows us to write where F_B represents the total buoyant force acting on the object, F_g represents the force of gravity acting on the object, and m represents the mass of the object. Since both the water and the oil contribute to the total buoyant force, we may write where

F_Bw

and

F_Bo

respectively represent the buoyant force of the water and oil. According to Archimedes' Principle, the buoyant acting on an object is equal to the weight of the fluid displaced. This allows us to write the following as the buoyant force of water. In this expression, each term represents the buoyant force of water acting on the object, however it is expressed in terms of different quantities. For example, the first two terms of the expression tell us that

F_Bw = F_{gw disp},

that is the buoyant force of the water is equal to the force of gravity acting on the water displaced (the weight of the water displaced). The third term expresses the weight of water displaced in terms of the mass of the water displaced. The fourth term expresses the mass of water displaced in terms of the mass density of water and the volume of water displaced, and the last term expresses the volume of water displaced in terms of the cross-sectional area of the object and the depth of the object in water. In like manner, we may develop an expression for the buoyant force of the oil. Here it should be pointed out that since we are interested in the depth of the object in water, we have used the fact that

h_{o disp} = h − h_{w disp}.

Combining the expression for the total buoyant force

(F_B = mg)

with the expression for the buoyant force of the water

(F_Bw = ρ_wd²h_{w disp}g)

and the buoyant force of the oil

(F_Bo = ρ_od²(h − h_{w disp})g),

we may write Simplifying and solving for the depth of the object in the water we obtain

Given the relationship between force and the electric field, We plug in our values and solve. Since Newton's second law is mass related to the force between two charged objects, Solving for acceleration gives us our desired result. Since the particle is positively charged it will experience a force in the direction the electric field points.