Newton's Law( Traffic Light Problem )

Lecture 10 Applications of Newton’s Laws and Friction When we are interested in the motion of an object, we have to identify ALL forces acting on that object. If the problem involves more than one object, then we have to identify all forces acting on each of the objects. The sketch of all forces acting on a body is called a free-body diagram. Since the corresponding reacting forces do not act on the object itself, but on other object, the free-body diagram does not include the reacting forces (it includes only the forces that act on the body and hence affect its motion). Sometimes objects are connected by strings. In that case the tension of the string is important. The tension is the force that tends to stretch the string. We will always assume that the string has no mass and that the tension is the same everywhere on the string. For example, if two persons are pulling the opposite side of a rope, in each point of the rope there is a pair of forces trying to stretch the rope, and this forces are the same everywhere on the rope: -T

T

-T

T x-axis

Objects in Equilibrium We say that an object is in equilibrium when it is at rest or it is moving at constant velocity, which means that an object in equilibrium has zero acceleration. According to Newton’s 2nd law, this means that the sum of all forces (or net force) acting on an object in equilibrium must be zero: ΣF=0 In a two-dimensional problem, we can separate this equation into its two component, in the x and y directions: Σ Fx = 0 Σ Fy = 0

Problems involving Newton’s laws in equilibrium and in motion. In Equilibrium Traffic light at rest (example 4.5 in the textbook) Traffic light weight: 1.00x102 N One cable attached to two other cables making angles of 37.0 and 53.0 degrees. What is the tension in each of the three cables? Solution: In this problem we must sketch two free-body diagrams, one for the traffic light, the other for the point of connection of the three cables.

Fig. 412, p.94

Since there are three unknowns (the tension in each of the three cables), we will need three equations. As the first equation, we can write Newton’s 2nd law for the traffic light. The traffic light has the force of gravity pointing down, and the tension of cable 3 pointing up, only vertical forces. Since the light is in equilibrium, we can write: Σ Fy = 0

⇒ T3 – Fg = 0

⇒ T3 = Fg = -1.00x102 N

We then apply Newton’s 2nd law to the point of connection of three cables, which is also in equilibrium, so the sum of all forces (the three tension forces) must be zero. However, this is now a two-dimensional problem, so the equation can be decomposed into its x and y components, giving two equations. This is great because we will a total of three equations, as needed. In the x direction we have: Σ Fx = 0

⇒ - T1 cos(37deg) + T2 cos(53deg) = 0

And in the y direction: Σ Fy = 0

⇒ T1 sin(37deg) + T2 sin(53deg) + T3 = 0

This is a system of two equations with two unknowns (T1 and T2), so it can be solved. Let’s first find T2 from the first equation: T2 = T1 cos(37deg) / cos(53deg) = T1 (0.799 / 0.602) = 1.33 T1 Then we substitute in the second equation this value of T2 and the value of T3 we had found earlier : T1 sin(37deg) + 1.33 T1 sin(53deg) + ( -1.00x102 N) = 0 Therefore: T1 0.60 + T1 1.06 – 100.00 = 0 ⇒ T1 = 100.00 / 1.66 = 60.241 = 60.2 N Where I have kept only 3 significant figures in the result of the final division, because that is the smaller number of significant figures of the two numbers I divide (3 significant figures in 1.66). Finally: T2 = 1.33 T1 = 80.066 = 80.1 N

In Motion Two hanging objects connected by a light string passing over a frictionless pulley (example 4.10 in the textbook) Objects masses: m1 and m2 What are the acceleration of the objects and the tension on the string? Solution: In this problem we need two free-body diagrams as well, one for each body.

On each body we have the tension of the string pulling up and the force of gravity pulling down. We have two unknowns (T and a) so we need two equations. We can simply write Newton’s 2nd law for each of the object. Sine this problem is one-dimensional (in the

vertical direction), each application of the 2nd law is a single equation, so two equation altogether, as we wanted: m1 a1 = T - m1 g m2 a2 = T – m2 g It looks like we have three unknowns, a1, a2 and T. However, it is clear that the two objects must have the same acceleration (though with opposite sign), because they are connected by the string, so we can write a third equation: a2 = – a1 So the second equation becomes: - m2 a1 = T – m2 g ⇒

m2 a1 = -T + m2 g

We then take the sum of this equation with the first equation and get: (m1 + m2) a1 = - m1 g + m2 g and therefore: a1 = g (m2 – m1) / (m1 + m2) Which shows that the acceleration of the object 1 is positive (upwards) if its mass is smaller than that of body 2, as one would have expected. We can then find the tension from the equation for object 1: m1 a1 = T - m1 g ⇒ m1 g (m2 – m1) / (m1 + m2) = T - m1 g ⇒ T = m1 g + m1 g (m2 – m1) / (m1 + m2) = 2 g (m1 m2) / (m1 + m2)

Problem Involving Friction Connected objects (example 4.13 in the textbook) Block mass: m1 = 4.00 kg Ball mass: m2 = 7.00 kg Coefficient of kinetic friction between block and surface: µK = 0.300. What are the acceleration of the objects and the tension on the string?

Fig. 418, p.99

First we will solve the problem considering the two objects separately (general method), then the whole thing as a single system (system approach). Solution with the general method: First of all, the magnitude of the accelerations of the two objects is the same, with positive x direction for the block and negative y direction for the ball. So we can write: a2 = - a1 We must sketch two free-body diagrams, one for the block and one for the ball. In the case of the block we have 4 forces: gravity (-m1g) and the reaction to it form the surface (n = m1g); the string tension T and the kinetic friction force fK . For the ball we only have the tension force T and the gravity force (-m2g).

Fig. 4p.1

We then write Newton’s 2nd law for the x-direction motion of the block and for the ydirection motion of the ball: T + fK = m1 a1 T – m2 g = m2 a2

⇒

T – m2 g = - m2 a1

The kinetic friction is equal to the normal force multiplied by the coefficient of kinetic friction, with direction opposite to the direction of motion, so we need a negative sign: fK = - µK n = - µK m1 g So the equation for the block becomes: T – µK m1 g = m1 a1

⇒

T = µK m1 g + m1 a1

And therefore the equation for the ball becomes:

µK m1 g + m1 a1 – m2 g = - m2 a1

⇒ a1 (m1 + m2) = g (m2 - µK m1)

⇒ a1 = g (m2 - µK m1) / (m1 + m2) = 5.17 m/s2 and T = µK m1 g + m1 g (m2 - µK m1) / (m1 + m2) = g (1 - µK) (m1 m2) / (m1 + m2) = 32.4 N

Solution for the acceleration using the system approach: Using the system approach we can view the whole thing as a single system, so we apply Newton’s second law to both objects together (because they are connected, so they must have the same acceleration): (m1 + m2) a = – m2 g + µK m1 g

⇒

a = - g (m2 - µK m1) / (m1 + m2) = - a1

which is the correct answer, because I have used the signs thinking of the motion in the y direction, so I have chosen a = a2 . But if you wanted to find again the tension, you would have had to use the two free-body diagrams, as the system approach can never give internal forces (forces internal to the systems). Furthermore, the system method can induce you to sign errors very easily, so you’d better avoid it!