in this video we're going to talk about how to draw free body diagrams so let's start with this problem let's draw the free body diagram for each of the following situations so for part a we have a box that is resting on a table how can we draw the free body diagram for that a free body diagram is a picture that shows all of the forces acting on an object so here we have a box what forces is acting on this box well we know that there's always going to be a weight force which we can call w the weight force is equal to the mass times the gravitational acceleration on earth g is 9.8 meters per second squared now the surface also exerts a force the surface exerts a force which is known as the normal force the normal force is always perpendicular to the surface it's at a 90 degree angle now because these two are equal because the box is at rest the normal force is going to be equal to mg now because they're equal you want the length of these two arrows to be approximately the same so that's it for part a this is the free body diagram that we can draw to represent the box resting on a table okay so now let's move on to part b a block attached to a rope hangs from a ceiling draw the free body diagram for that so let's say this is the ceiling and we're going to have a block that is attached to one rope that connects the block to the ceiling so what forces are acting on this block so we have the weight force which we can call w or you can say f gravity and we also have an upward tension force now this block is at rest it's not accelerating upward or downward so the net force in the y direction is zero therefore these two arrows should be of the same left so in this example the tension is equal to the weight force and the weight force is mg so the tension in the rope is simply mg tension is always the force that acts through a rope now what about part c a block is pulled upward at constant velocity to pull a block upward we need a rope so the block is moving upward at constant velocity we're going to have a weight force and we're going to have an upward tension force are these two forces the same or are they different at constant velocity what is the acceleration the acceleration is zero at constant velocity the net force in the y direction is going to be the difference between a tension and the weight force and according to newton's second law the net force which is the sum of the forces it's going to equal to m times eight so we can replace this with m a now if the acceleration is zero the net force is going to be zero adding w to both sides we can see that the wave force and the tension force they're the same so by constant velocity these two forces will cancel out because the acceleration is zero so therefore the length of the two arrows should be the same so that's the tension in this problem is equal to the weight force which is equal to mg so that's all we need to do to show the free body diagram for part c now let's move on to part d this time a block is pulled upward with a constant acceleration so the equation changes so we can still write this equation the sum of the forces in the y direction that's still going to be equal to tension minus the weight force and the net force in the y direction is going to be m a you can put a sub y if you want to for acceleration in y direction so this is tension minus mg if we move mg to the other side notice that we have m a plus mg so that's how you can calculate the tension for this example if there's an upward acceleration so notice that tension is greater than mg it's mg plus ma therefore this arrow has to be longer than the arrow for the wave force or f gravity so let's make this arrow significantly bigger than the other one so if t is bigger than w there's going to be an upward net force that's going to accelerate the block upward now what about e a block let's take this word out a block descends lower with a downward acceleration using a rope so in this case the acceleration is negative if it's going down the net force is going to be in the downward direction because it's descending lower with a downward acceleration and the only way that's going to happen is if w is greater than t so we're going to make this arrow smaller we're going to make this one bigger so now we have a downward acceleration because the wave force or the force of gravity is greater than the tension force so that's it for e and that's the free body diagram that correlates to that situation so now let's work on some more examples for part a we have a block that slides across a frictionless horizontal surface at constant speed go ahead and draw a free body diagram for that situation so first we have a horizontal surface and then we have a block on that surface and let's say it's moving to the right at constant speed what forces are acting on this block well any time you have an object resting on a surface there's going to be an upward normal force and that upward normal force is going to be equal to the weight force so we need to make sure the lengths of those two arrows are the same now are there any forces in the x direction if the object is moving at constant velocity the net force or the sum of the forces in the x direction will be zero there's no friction slowing this down so we don't have friction therefore there's no applied force accelerating it to the right so there's no forces in the x direction here so this is all we could show for the free body diagram for part a now let's move on to part b so in part b we have an applied force that is used to push the box to the right at constant speed so this is going to be the applied force now we still have a num our normal force so that's going to be present and the wave force is still there now the box is moving at constant speed if it's moving at constant speed the acceleration is zero so the force is in the x direction must be balance because the net force is zero in the x direction so therefore if the net force in the x direction must be zero what other forces do we have the only other force that can be acting in the x-direction must be friction now there's two types of frictional forces there's static friction and kinetic friction the word kinetic indicates that the object is in motion and since this object is moving at constant speed kinetic friction is active which we'll call fk now these two forces must be equal in order for the acceleration to be zero so the arrow these two arrows have to have the same length so that's the answer for part b part c an applied force is used to accelerate a box across a horizontal surface with friction present so friction was already present in part b but this time the box is not moving at constant speed it's accelerated so the acceleration is not zero because we have an acceleration there's going to be a net force to the right and the only way that's going to happen is if the applied force is greater than fk so kinetic friction has to be relatively small compared to the applied force so that we can have a net acceleration to the right so this is going to be the free body diagram for part c now what about part d a rope is used to pull a block to the right across a horizontal surface with friction present at constant velocity so the normal force and the weight force will still be the same for part d we no longer have an acceleration we're moving at constant velocity so the forces in the x direction must be balanced this time we're using a rope to pull the block to the right so that's gonna be our tension force and that's gonna the block is being pulled to the right against friction friction always opposes motion so if we're moving to the right friction is going to be directed to the left so in this case fk it's going to be over here and it's going to have the same length as the tension force vector so this is the free body diagram that corresponds to situation the situation part d now let's move on to part e a rope directed at 30 degrees above the horizontal is used to pull a block to the right across a horizontal surface with friction at constant acceleration so we no longer have constant velocity we're going to accelerate the block to the right so the forces in the x direction will not be equal now if we're pulling the block to the right friction will still be directed towards the left but if we're pulling it to the right with an acceleration friction is going to be less than the tension force so we're going to draw a relatively small fk value to the left and a large tension force to the right but at an angle so that tension force has an x component and it has a y component it's directed at an angle of 30 degrees above the horizontal the x component of the tension force is t x and t y is the y component in order to accelerate the block to the right tx has to be greater than fk now in other problems the normal force was equal to mg but that's not going to be the case in this example because of the y component of t the block is not accelerating upward or downward so the net force in the y direction is zero and it's equal to these two upward forces which is positive fn plus positive t y and then we have a downward force so this is going to be negative w now because the block is not accelerating upward or downward as was mentioned before the net force in the y direction is zero so if we take these two terms move it to the other side we'll get that the normal force is equal to t y is actually going to be w positive w minus t y so the normal force is not mg in this problem the normal force is the way force mg but minus the y component of the tension force so that's how you can calculate the normal force for this particular situation now what about calculating the acceleration of the block in the x direction how can we come up with a formula that's going to help us to get that answer before we do that let's talk about how to get t y and t x using sohcahtoa and trigonometry t y is equal to t sine theta t x is equal to t cosine theta so going back to trig we have the expression sohcahtoa the so part means sine is equal to the opposite side over the hypotenuse so sine of the angle theta theta is 30 degrees in this example is equal to opposite which is t y divided by the hypotenuse which is t the hypotenuse is across the right angle it's the longest side of the triangle and so if you multiply both sides by t you get t y is equal to t sine theta for the other equation we have cosine theta so this is the co part cosine is equal to the adjacent side divided by the hypotenuse which is t so multiplying both sides by t you get t x is equal to t cosine theta so that's how you can get t x and t y in this problem if you know t so you can plug t sine theta into this formula to get the normal force but now to get the acceleration in the x direction we need to start with this expression the sum of the forces in the x direction is t x it's positive because it's going to the right this is negative because it's going to the left so it's t x minus f k now the net force which is the sum of the forces is always equal to m a based on newton's second law so this is going to be m a sub x acceleration in the x direction t x is t cosine theta now f k is equal to mu k times the normal force in this example remember the normal force is not just mg it's mg minus ty so i'm going to leave it like this so first you want to calculate tx and ty once you have that calculate the normal force when you have the normal force you have everything you need to calculate the acceleration in that equation so that's how we can find the acceleration in the x direction it's by using that formula now let's focus on free body diagrams with inclines so part a a block slides down a frictionless incline let's draw a picture so here is our incline and here is the box on the incline what forces are acting on this box so we have a downward weight force we can just call that w and we have a normal force the normal force is less than the weight force so that arrow should be smaller than w the extent to which like how much smaller depends on the angle of the incline as the angle of the incline increases fn becomes much smaller relative to w so on an incline the normal force is going to be mg cosine theta as the angle increases from 0 to 90 cosine decreases from 1 to 0. so as theta goes up from 0 to 90 the normal force goes down so the length of this arrow will get smaller as theta increases now what other forces do we have acting on this picture or the block now we have a component of gravity that will accelerate the block down the incline and i'm going to call that fg fg is equal to mg sine theta so as the block slides down on an incline that does not have friction it's not going down at constant speed it's accelerating downward the only way you can go down at constant speed is if friction is present and if it's equal to fg but because friction is not present this is the only force acting parallel to the incline so there's an acceleration to calculate that acceleration let's define this as the y-axis and this has the x-axis with respect to the block so the sum of the forces in the x-direction is equal to fg that is the only force acting along the x-axis parallel to the incline and this is always equal to m a f g is mg sine theta so we could divide both sides by m so those variables will cancel thus the acceleration for this problem part a is simply equal to g sine theta so that's how you can calculate the acceleration of a block that's sliding down a frictionless incline now for those of you who might be wondering how we can get these equations here's what you can do so draw a line that is parallel to the normal force we're going to create a triangle where w is the hypotenuse of that triangle so let me just put this in a different color so you can see it so there's the triangle and here's the right angle and this angle theta is equivalent to this angle so across theta that's the opposite side we know that sine is opposite over hypotenuse so i'm going to redraw this triangle so the hypotenuse is the wave force which is mg this is theta across theta this is going to be mg sine theta and on the adjacent side cosine is associated with the adjacent side so this side here that's going to be mg cosine theta so notice that fg is equal to this component of gravity because that component is parallel to fg so that's why fg is equal to mg sine theta the normal force balances this portion of the triangle so that's why the normal force is mg cosine theta so that's how you can get those equations from that free body diagram now part b a block remains at rest on an incline so the block is still on the incline we're still going to have the normal force the wave force and fg those forces will always be present for an object on an incline but this time in part b the block is not sliding down the frictionless incline it's remaining at rest the only way it can remain at rest is if the net force in the x direction is zero so there must be some other force that's preventing it from sliding down and that force must be equal in magnitude to fg so what is that force well that can only be friction friction opposes motion remember there's two types of frictional forces one of you whenever you have objects like sliding past each other you have kinetic friction when they're sliding past each other and static friction when they're at rest so because the block is at rest static friction is the force that is going against fg that's preventing it from sliding down so in this example fs is equal to fg now the static frictional force is less than or equal to mu s times the normal force so static friction is an inequality it's going to match fg up to a maximum so let's say that maximum is a hundred if we were to make a table that shows the relationship between fg and fs if fg is zero fs is zero if fg is 20 fs is 20. if fg is 60 fs is going to match it to 60. if fg is 100 fs will be 100. if fg is 110 static friction has been exceeded the block will begin to slide down and you're not going to have static friction anymore you're going to have kinetic friction so static friction will match fg up to a certain limit at which point the block will begin to slide down but this is the free body diagram for part b now part c a block slides down an incline with friction present at constant acceleration so let's delete this so there's constant acceleration which means fg has to be greater than friction and because the block is sliding down we no longer have static friction but we have kinetic friction so fk is going to be smaller than fg now how can we calculate this acceleration well we can write this equation the sum of the forces in the x direction is going to be f g minus f k the net force in the x direction will always equal m a we know f g is mg sine theta f k that's mu k times the normal force by the way static friction fs is mu s times the normal force but notice that it's an inequality it's going to match up it's going to match fg up to a certain limit fk does not have that inequality it's simply equal to mu k times the normal force so this is a fixed value it doesn't depend on the value of fg so make sure you're aware of that difference in between fs and fk so fk is going to be mu k times the normal force and the normal force is mg cosine theta so we can just replace fn with mg cosine beta since each term contains m we can divide every term by m therefore the acceleration in the x direction for this example is going to be g sine theta minus mu k g cosine theta so that's how you can calculate it for this problem that is for part c now let's move on to part d a rope is used to pull a block up an incline against friction at constant velocity feel free to try that problem so let's draw a different incline this time we're going to draw an incline that goes up as the block moves to the right so let's put the block here and we're going to pull the block up an incline so this is going to be our tension force we still have our downward weight force and we're still going to have the normal force which is going to be less than the weight force as long as there's an angle the only way they'll be equal if the angle is zero so the angle is not zero for an incline thus normal force has to be less than the weight force now we're pulling it up against friction at constant velocity so because it's moving we're dealing with kinetic friction and because the velocity is constant there's no acceleration so the forces are balanced in the x-direction which means that tension force is equal to fk so if we wish to calculate the tension force in this case we could say the sum of the forces in the x direction is equal to positive t minus f k now because the acceleration is zero and this equals m a m times zero is zero so this is zero adding f k to both sides we get that f k is equal to t where t is equal to f k and f k it's mu k times the normal force and the normal force on the incline is mg cosine theta so for this specific problem where we're pulling the block up an incline against friction at constant velocity the tension force is mu k mg cosine theta you