Chapter 4 Lecture 2

Statics

Statics is the study of objects in equilibrium. An object is said to be in equilibrium (translational) if it is either at rest or moving in a straight line with a constant speed. According to Newton's first law, this happens when the net force acting on the object is zero.

Condition of Equilibrium

The condition of equilibrium states that an object will be in equilibrium (translational) if the sum of all the forces acting on the object is zero.

F_net = Σ F = F₁ + F₂ + . . . = 0

where F₁ , F₂ , . . . are all the forces acting on the object. This is a vector equation. It can be decomposed into an equation for the horizontal components and an equation for the vertical components to give two algebraic equations. If a vector is equal to zero, then its components are also equal to zero.

Σ F_x = F_1x + F_2x + . . . = 0

Σ F_y = F_1y + F_2y + . . . = 0

where F_1x + F_2x + . . . and F_1y + F_2y + . . . are the horizontal and vertical components of all the forces acting on an object in equilibrium respectively.

Solving Equilibrium Problems

Identify all the forces acting on the object. Identifying a force most of the time means identifying the magnitude and the direction of the force. In identifying direction, it is a good practice to obtain angles measured from the default reference line (east or positive x-axis). Representing the force vectors graphically helps to visualize the problem better. The contact forces can easily be identified by identifying the objects in contact with the object under consideration. There is always the non-contact force weight. The direction of weight is downwards or towards the ground (its default angle is -90°). Its horizontal component is zero. Its vertical component is negative because it points downward.

After all the forces have been identified (along with the unknowns), Find the horizontal and vertical components of each force with the appropriate signs. If the default angle (angle with respect to the positive x-axis) is used, the formulas for calculating components will yield the appropriate sign. If angles with respect to other reference lines are used, the appropriate sign should be included according to the sign conventions for components. Magnitude is always positive. Plug in the components into the two equations of condition of equilibrium in the form of components and solve for the unknowns. These equations can be used to solve for two unknowns.

Example: A 10 kg object is hanging from a ceiling by means of a string. Calculate the tension in the string.

Solution: There are two forces acting on this object in equilibrium: its weight and the tension in the string. The drection of the force due to the tension in the string is upwards. Since both vectors are vertical only the equation for vertical components need to be used. Tension in the string will be represented by T.

m = 10 kg; T = ?; θ_T = 90° ; θ_w = -90° ; T = ?

w = m|g| = 10*9.8 N = 98 N

w_y = w cos θ_w = 98*sin (-90°) = -98

T_y = T sin θ = T sin (90°) = T

T_y + w_y = 0

T = T_y = -w_y = -(-98) N = 98 N

Example: A 100 kg object is hanging from a string by means of two strings. One of the strings is directed at 37° north of east. The other string is directed at 53° north of west. Calculate the tensions in the strings.

Solution: There are three forces acting on this object in equilibrium: its weight and the tensions in the two strings.

m = 100 kg; θ_T₁ = 37° ; θ_T₂ = (180 - 53)° = 127°; T₁ = ? ; T₂ = ?

w = m|g| = 100*9.8 N = 980 N

w_x = w cos θ_w = 980 cos -90° N = 0 N

w_y = w sin θ_w = 980 sin -90° N = -980 N

T_1x = T₁ cos θ_T₁ = T₁ cos 37° = 0.8T₁

T_1y = T₁ sin θ_T₁ = T₁ sin 37° = 0.6T₁

T_2x = T₂ cos θ_T₂ = T₂ cos 127° = -0.6T₂

T_2y = T₂ sin θ_T₂ = T₂ sin 127° = 0.8T₁

T_1x + T_2x + w_x = 0.8T₁ - 0.6T₂ + 0 N =0 N

T₁ = 0.75T₂

T_1y + T_2y + w_y = 0.6T₁ + 0.8T₂ + -980 N =0 N

0.6(0.75T₂) + 0.8T₂ = 980 N

T₂ = 980/(0.6*0.75 + 0.8) N = 784 N

T₁ = 0.75T₂ = 0.75*784 N = 588 N

Dynamics

Dynamics is the study of accelerated systems. According to Newton's second law, the product of the mass of an object and its acceleration is equal to the net force acting on the object.

F_net = Σ F = F₁ + F₂ + . . . = ma

where F₁ + F₂ + . . . are all the forces acting on an object of mass m moving with an acceleration of a. This is a vector equation. It can be decomposed into two one dimensional algebraic equations: one relating the horizontal components and the other relating the vertical components. This is possible because the horizontal and verical motions are independent in a sense that acceleration in one doesn't affect motion in the other.

Σ F_x = F_1x + F_2x + . . . = ma_x

Σ F_y = F_1y + F_2y + . . . = ma_y

where F_1x + F_2x + . . . and F_1y + F_2y + . . . are the horizontal and vertical components of all the forces acting on the object respectively.

Solving Dynamics problems

Identify all the forces acting on the object. The non-contact force, weight, is always there. The contact forces can easily be determined by looking at the objects in contact with the object. If the object is sliding on a surface, make sure to include the force exerted by the surface. The vertical component (normal) and the horizontal component (friction) of a surface force are customarily dealt with separately. The normal force is perpendiclar to both surfaces and is directed towards the surface of the object under consideration. If the object is moving in a straight path, the direction of friction is parallel to the surfaces and opposes the direction of motion.

Find the horizontal and vertical components of all the forces (including unknown forces). In calcuating components, it is a good idea to use the angle measred with respect to the positive x-axis so that you don't have to worry about having the right sign for the components. Magnitude is always positive. Weight is directed towards the ground. Its horizontal component is zero. Its vertical component is negative and is numerically equal to the weight. Its default angle is -90°. For an object sliding on a horizontal surface, the default angle for the normal force is 90°. Its horizontal component is zero. Its vertical component is positive and is equal to the magnitude of the normal force. Again, if the object is sliding in a straight path, the direction of friction is parallel to the surfaces and opposes the direction of motion. Its default angle is 180° (assuming the motion is along the positive x-axis). Its vertical component is zero. Its horizontal component is negative and is numerically equal to the magnitude of friction. Most of the time, you will be dealing with horizontal motion in a straight line. In such cases, the vertical component of acceleration is zero and the horizontal component is numerically equal to the magnitude of acceleration.

Example: An object of mass 20 kg is moving on a frictionless horizontal surface. It is being pulled by a string that makes an angle of 60° with the horizontal. The tension in the string is 100 N. Calculate the acceleration of the object.

Solution: The forces acting on the object are its weight, the tension in the string and the force exerted by the surface in which it is sliding. Since the surface is frictionless, the surface force consists of the normal force only. The vertical component of the acceleration is zero because it is moving on a horizontal surface. The only force with a horizontal component is the tension in the string which is already known. Thus, only the horizontal component of Newton's law needs to be considered.

m = 20 kg ; T = 100 N ; θ_T = 60° ; a = ?

a_x = a

T_x = T cos θ_T = (100 cos 60°) N = 50 N

T_x = ma_x = ma

a = T_x ⁄ m = 50 ⁄ 20 m ⁄ s² = 2.5 m ⁄ s²

Example: A 10 kg object is moving in a frictionless horizontal surface. It is being pulled by a 20 N force that makes angle of 37° with the horizontal. It is also being pulled backwards by a horizontal force of 5 N. Calculate the acceleration of the object.

Solution: The forces acting on this object are its weight, the normal force exerted by the surface (friction is zero because the surface is frictionless) and the tensions in the two strings. The acceleration is horizontal because it is moving in a horizontal surface. The only forces with horizontal components are the tensions in the strings. Hence, only the horizontal component of Newton's second law needs to be used.

m = 10 kg ; T₁ = 20 N ; θ_T₁ = 37° ; T₂ = 5 N ; θ_T₂ = 180°

a = a_x

T_1x = T₁ cos θ_T₁ = (20 cos 37°) N = 16 N

T_2x = T₂ cos θ_T₂ = (5 cos 180°) N = -5 N

T_1x + T_2x = ma_x = ma

a = (T_1x + T_2x) ⁄ m ={(16 + (-5)) ⁄ 10} m ⁄ s² = 1.1 m ⁄ s²

Example: A 10 kg object is being pulled by a string that makes an angle of 30° with the horizontal. The tension in the string is 100 N. The coefficient of kinetic friction between the surfaces is 0.2.

Calculate the normal force exerted by the surface of the ground on the object.

Solution: The forces acting on the object are its weight, surface force which includes both friction and normal force and the tension in the string. The normal force is a vertical force directed upward. It has only a verical component which is equal to the magnitude of the normal force. The other forces with vertical components are its weight and the tension in the string which are already known. Thus, only the vertical component of Newton's second law can be used to calculate the normal force.

m = 10 kg ; T = 100 N ; θ_T = 30° ; N = ?
N_y = N

w_y = -w = -m|g| = -10*9.8 N = -98 N

T_y = T sin θ_T = (100 sin 30°) N = 50 N

N_y + w_y + T_y = 0 N

N = N_y = -(w_y + T_y ) = -(-98 + 50) N = 48 N

Calculate the force of friction.

Solution: μ = 0.2 ; f = ?

f = μN = 0.2*48 N = 9.6 N

Calculate its acceleration.

Solution: The acceleration has only a horizontal component which is equal to the magnitude of acceleration (assuming it is going to the right) because it is moving horizontally. The forces with horizontal components are friction, and the tension in the string. The angle for friction is 180° because it opposes the motion.

θ_f = 180° ; a = ?
a_x = a

f_x = f cos θ_f = (9.6 cos 180°) N = -9.6 N

T_x = T cos θ_T = (100 cos 30° ) N = 87 N

T_x + f_x = ma_x = ma

a = (T_x + f_x ) ⁄ m = {(87 + (-9.6)) ⁄ 10 } m ⁄ s² = 7.74 m ⁄ s²

The Inclined Plane

Consider an object of mass m sliding in an inclined plane whose inclination with respect to the horizontal is θ. The force responsible for the sliding motion of the object is the component of gravitational force along the plane. Since gravitational force is vertical, the angle between the plane and the direction of its weight is (90 - θ). The sine of an angle is equal to the cosine of its complementary angle. The cosine of an angle is equal to the sine of its complementary angle. Thus, the component of gravitational force parallel to the plane is given by sine and the component perpendicular to the plane is given by cosine.

w_|| = m|g| sin θ

w_⊥ = m|g| cos θ

w_|| is the component of gravitational force parallel to the plane. w_⊥ is the component of gravity perpendicular to the plane. If the inclined plane is frictionless, the aceleration down the plane can be obtained by dividing the component of its weight along the plane by its mass.

a = (m|g| sin θ ) ⁄ m = |g| sin θ

a is the acceleration of the object down the plane when friction can be neglected.

Example: A 10 kg object is sliding down a 30° inclined plane.

Calculate the component of gravitational force down the plane.

Solution: m = 10 kg; θ = 30 ° ; w_|| = ?
w_|| = m|g| sin θ = (10*9.8 sin 30° ) N = 49 N

Calculate the component of gravitational force perpendicular to the plane.

Solution: w_⊥ = ?

w_⊥ = m|g| cos θ = (10*9.8 cos 30° ) N= 85.3 N

Calculate its acceleration down the plane.

Solution: a_|| = ?
a_|| =|g| sin θ = (9.8 sin 30° ) m ⁄ s² = 4.9 m ⁄ s²

Systems Involving more than One Objects

Newton's second law can be applied provided only external forces are considered. External forces are forces exerted on the system by objects outside the system. A force exerted by one of the objects in the system on another object in the system is an internal force and should not be considered.

Σ F_ext = m_totala

Example: A 2 kg object is connected with an 8 kg object on a frictionless surface by means of a string. The 8 kg object is being pulled horizontally by a force of 100 N. Calculate the acceleration of the system.

Solution: F_ext = 100 N ; m_total = (2 + 8 ) kg = 10 kg ; a = ?

F_ext = m_totala

a = F_ext ⁄ m_total = 100/10 m ⁄ s² = 10 m ⁄ s²