A box of mass m slides across a horizontal table with coefficient of friction μ. The box is connected by a rope which passes over a frictionless pulley to a body of mass M hanging along side the table. Find the acceleration of the system and the tension in the rope.

Solution:

- Concepts:

Newton's second law - Reasoning:

For each box we have**F**= m**a**. The magnitude of**a**is the same for each box. - Details of the calculation:

Mg – T = Ma.

T – f = ma.

f = μmg.

a = (M – μm)g/(M + m), T = Mg – Ma = Mmg(1 + μ)/(m + M).

In the figure below the coefficient of friction is the same at the top and the
bottom of the 700-g block.

(a) Draw a free body diagrams for both the 200-g and the 700-g blocks,
considering all forces.

(b) If the acceleration is a = 70 cm/s^{2} when F = 1.3 N, how large is
the coefficient of friction?

Solution:

- Concepts:

Newton's second law - Reasoning:

This is a simple Newtonian Mechanics problem. - Details of the calculation:

(a) Forces acting in the horizontal direction:

(b) F – T – f_{1}– f_{2}= Ma, T – f_{2}= ma.

f_{1}= μ(M + m)g, f_{2}= μmg.

F – ma – f_{1}– 2f_{2}= Ma.

F – μ Mg – 3μ mg = (M + m)a.

μ = [F – (M + m)a]/[Mg + 3mg].

μ = [1.3 – 0.9*0.7]/[1.3*9.8] = 0.0526 .

A trunk weighing 500 N is to be pushed up a rough incline by an applied **
horizontal** force **F**. The incline makes an angle of 40^{o} with the
horizontal. If a force of 1000 N is sufficient to move the trunk at a constant
velocity of 0.200 m/s, what is the coefficient of kinetic friction between the
trunk and the incline?

Solution:

- Concepts:

Newton's 2^{nd}law, the force of friction, the normal force, the gravitational force - Reasoning:

The trunk is moving in an inertial frame with constant velocity. It is not accelerating, Newton's 2^{nd}law requires that the net force is zero. In addition to the applied force, the force of friction, the normal force, and the gravitational force are acting on the trunk

The vector sum of all the forces acting on the trunk must be zero. - Details of the calculation:

Choose the axes of your coordinate system as shown. The x- and y-components of the net force must be zero.

x-component:

f + mg sin(40^{o}) - F_{a}cos(40^{o}) = 0. Solve for f.

f = (1000 N) cos(40^{o}) - (500 N) sin(40^{o}) = (444.65 N).

y-component:

N – mg cos(40^{o}) – F_{a}sin(40^{o}) = 0. Solve for N.

N = mg cos(40^{o}) + F_{a}sin(40^{o}) = (1025.81 N).

Use f = μN to find the coefficient μ.

μ = f/N = 0.43.

A block is given a quick push along a horizontal table. The coefficient
of kinetic friction between the block and the table is μ_{k}. It
is known that during the time interval t (immediately after the push) the block
covers a distance d. Find the distance that may be covered by the block
during the subsequent time interval t'. Find all possible answers.

Solution:

- Concepts:

Newton's 2^{nd}law, friction, kinematics - Reasoning:

The force of kinetic friction is responsible for the acceleration of the block. - Details of the calculation:

Let the block's mass be m, its momentum just after it received the impulse be**p**= mv_{0}**i**.

dp_{x}/dt = -μ_{k}mg until the block comes to rest.

Therefore p_{x}decreases at a steady rate until the block comes to rest, a_{x}= μ_{k}g.

The block comes to rest at time t_{rest}= v_{0}/(μ_{k}g).

The block can cover a maximum distance d_{max}= v_{0}t_{rest}-_{ }½μ_{k}gt_{rest}^{2}= ½v_{0}^{2}/(μ_{k}g).

Case (a) t ≥ t_{rest}, d = d_{max}, then the distance covered during the subsequent time interval t' is zero.

Case (b) t + t' ≥ t_{rest}, then the distance covered during the subsequent time interval t' is d_{max}- d.

Case (c) t + t' < t_{rest}, then the distance covered during the subsequent time interval t' is

d' = v_{0}(t + t') -_{ }½ μ_{k}g(t + t')^{2}– d = v_{0}t' -_{ }½ μ_{k}g t'^{2}- μ_{k}gtt'.

You are going to entertain children by pulling a tablecloth out from under
the the cake at a birthday party. The birthday cake is resting on a
tablecloth at the center of a square table. Each side of the table has
length l = 1.84 m. The tablecloth is the same size as the table top.
You grab the edge of the tablecloth and pull sharply in a direction along one of
the sides. The tablecloth and cake are in contact for time t after you
start pulling. Then the sliding cake is stopped (you hope) by the friction
between the cake and table top. The coefficient of kinetic friction
between the cake and tablecloth is μ_{k1} = 0.310 , and that between the
cake and table top is μ_{k2} = 0.410. Calculate the maximum value
of t if the cake is not to end up on the floor. Neglect the size of the
cake compared to the size of the table.

Solution:

- Concepts:

Newton's 2^{nd}laws, kinematics - Reasoning:

Let the direction of motion of the tablecloth be the positive x-direction.

Assume that the cake moves a distance d while still on the tablecloth and a distance Δx while sliding on the table top.

Assume it falls off the table when its center reaches the edge of the table, i.e. when d + Δx = l/2. - Details of the calculation:

(i) The cake is still in contact with the tablecloth and is moving in the direction of motion of the tablecloth.

The only force acting on the cake in the horizontal direction is friction.

f = ma = μ_{k1}mg, a = μ_{k1}g in the positive x-direction..

d = ½μ_{k1}gt^{2}is distance the cake slides during its contact with the tablecloth.

Its velocity when it looses contact with the tablecloth is μ_{k1}gt, in the positive x-direction.

(ii) The cake is now in contact with the table.

Again, the only force acting on the cake in the horizontal direction is friction.

f = ma = -μ_{k2}mg, a = -μ_{k2}g. The direction of the acceleration is now in the negative x-direction.

The cake will move a distance Δx before it comes to rest.

v_{xf}^{2 }= v_{xi}^{2}+ 2aΔx. Δx = v_{xi}^{2}/(2μ_{k2}g) = (μ_{k1}gt)^{2}/(2μ_{k2}g).

To solve for the maximum t we set d + Δx = l/2.

½μ_{k1}gt_{max}^{2}+ (μ_{k1}gt_{max})^{2}/(2μ_{k2}g) = l/2.

t_{max}^{2}= l/[g(μ_{k1}+ μ_{k1}^{2}/μ_{k2})] = 0.345 s2.

t_{max}= 0.587 s.