Scalar, Vector & Static Equilibrium | Classical Mechanics | A-level Physics

Transcript

in this video we'll be discussing scalers vectors and static equilibriums let's begin with the scalar quantity we can Define Scala as a quantity that has magnitude or size but without direction to give a few examples we have distance speed mass energy temperature and volume we represent scalar quantities with a numerical value followed by its unit let's take a book as an example with two quantities mass and tempature firstly does it make sense to associate these quantities to the book well the book does in fact have mass and a temperature so it does make sense now ask yourself does it make sense to put a Direction with these quantities for example the mass of the book is 5 kg to the right or the temperature of the book is 15° C to the left well that doesn't really make sense so in a sentence if we put a direction to a quantity and it doesn't seem to make sense we can use that as a form of testing to see whether or not the quantity is in fact Scala moving on to Vector quantities we can define a vector as a quantity that has both magnitude and a direction to give a few examples we have displacement velocity acceleration force and weight let's represent a vector quantity by considering the forces that are acting on a ball here is an example we can see the force F1 of 5 Newtons acting on a ball the direction is represented by the arrowhead and the magnitude by the length so the ball is currently moving to the right say a force FS2 of 10 Newtons represented in a red arrow is acting on the same board but in the opposite direction taking the right hand side as a positive the resultant force on the ball would be 5 Newtons - 10 NT which would give - 5 Newt or 5 Newtons to the left let's do another example we have forces F3 of 10 Newtons to the right and F4 of 5 Newtons to the left what do you think the resultant force would be taking the right hand side as a positive the resultant force on the ball would be 10 Newt minus 5 Newtons which would give positive 5 Newtons or 5 Newtons to the right unlike quantities where the magnitudes can just be added together to get a definitive answer it is important to take direction of the forces into account when working with vectors if say the force on the object is acting in the opposite direction the forces should be subtracted as we saw on the example vectors can be resolved into two components using trigonometry the two components being horizontal and vertical an example of this is a ball moving at a a constant velocity we can draw the axes taking the right hand side and up as positive the ball is moving at a constant velocity of V m/s at an angle of theta from the horizontal we want to resolve the force into its X and Y components which are represented by the Red Arrows if I were to put the Red Arrows together head to tail you can see it forms a right angle triangle where the hypotenuse is the initial vector quantity V from from here we can use SOA to find three expressions for the angle tan Theta equal V y/ VX cos Theta = VX / V sin Theta = v y / V using cos and S we can calculate the horizontal and vertical components given the angle Theta and the resultant vector v rule of thumb use the cosine expression when resolving the vector through the angle and use S when resolving the vector away from the angle let's move on to calculating the resultant Vector if two perpendicular Vector components are given the resultant Vector is calculated using Pythagoras's Theorem we can draw the vector v with its X and Y components for example the resultant velocity is given by V ^2 = vx2 + v y squ the resultant Vector can also be calculated using trigonometry if one of the components and the angle is known you can refer to the equations derived using SOA TOA in a static equilibrium the resultant force acting on a body is zero therefore all the forces along the line of action must balance let's look at a few examples in the first example we have a light inextensible string here we can make two assumptions by light we mean that the string has no mass and by an extensible we mean that it does not stretch so we have a fixed Point like a ceiling a ball is attached to the ceiling using a light and extensible string in this case there are two forces acting on the ball the weight of the ball and the tension on the string since the ball is in step equilibrium the resultant Force must be zero this means that the two opposing forces should be equal now what if we had more than one string in example two we have a ball attached to two fixed points A and B using two light inextensible strings firstly we can mark the weight then we have the tension forces acting along the two strings at an angle Theta from the horizontal in this case since the tension is distributed along two strings it must be resolved vertically referring back to SOA we can see that the vertical component can be found using T sin Theta as the tension is distributed along two strings it is calculated twice this gives us 2 T sin Theta giving us the final equation 2T sins mg for the third example we will be working with a smooth slope here we can make an assumption that on any smooth surface there are no frictional forces exerted here we have a smooth slope at an angle Theta from the horizontal on the slope we have a ball attached to a box using a light and extensible string the mass of the ball is M and the mass of the Box capital M we can label the weight of each object and then the tension in the string as the ball is touching the slope there will also be normal contact force perpendicular to the slope how do we find the relationship between M and capital M firstly we need to resolve the weight of the ball into its horizontal and vertical components to the slope when we resolve it we get mg cos Theta as a vertical component and mg sin Theta as the horizontal represented in purple arrows now that we have resolved the forces acting on the board we can consider them parallell and perpendicularly to the slope let's begin with the parallel forces we can see that the only forces acting on the ball is the tension and the horizontal component of the weight note that the angle of the slope and the angle between the weight vector and its vertical component to the slope they are both Theta as the triangles are similar this gives T equals mg sin Theta taking perpendicular forces we get n equal to the vertical component of the weight mg cos Theta you can pause the video and check to see if you understand now consider the forces on the box here we don't need to resolve anything as the forces acting on it are vertical only therefore t equal mg we can assume the tensions on the string on both sides are equal as they are essentially just one light and extensible string with that in mind we can equate equations 1 and three to produce a new expression for M and capital M we can now find the ratio between the mass of the ball and the Box in this case the ratio between the two masses would be sin Theta in the final example we will be working on a rough slope a rough surface exerts a frictional force on the object one thing to note is that friction always acts oppositely to the direction of motion that the object would be subjected to so let's make this clearer here we have a rough slope at an angle Theta from the horizontal there is a box of mass m on the slope naturally the weight of the box would pull the box down the slope thus making the frictional force act upwards parallel to the slope we also have the normal contact force of the Box okay we want to find the expression for the coefficient of friction denoted by the Greek letter mu just like example three we need to resolve the weight of the box into its horizontal and vertical components to the slope the vertical component comes to mg cos Theta and the horizontal mg sin Theta in this equation the frictional force is Mu multiplied by n where mu is the coefficient of friction and n the normal contact force this doesn't appear on some of the physics syllabuses but it is a general equation for friction equations 2 and three are what you get when you resolve horizontally and vertically by equating equations 1 and 2 we get our fourth equation mu n = mg sin Theta Now by substituting equations 3 into four we get mu mg cos Theta = mg sin Theta we can now divide both sides by mg cos Theta to get mu = sin Theta cos Theta or just tan Theta which is the ratio between sin Theta and cos Theta so the value of mu can be determined if you know the angle of the slope that is to say the friction acting on the box depends on the angle the slope is making to the horizontal well that's it for this video if you have any questions or you are unsure about something you can leave a comment below thank you for watching please like And subscribe for more