Tutor profile: Eleanor J.
Subject: Physics (Newtonian Mechanics)
Consider a 3 lb bar of length 6 feet which is supported by 2 simple pin supports, they are 2.5 and 3.2 feet from the left end respectively. The bar has a 7 lb load 5.4 feet from the left end. Determine the forces applied on the bar by the two supports if the bar remains static.
The first step to every physics problem is to draw a free body diagram of the object which we are discussing. While I can't really draw much in this format I'll try my best. F1 w F2 L L = 7 lb down F2 = ? up? i_____________i____I__i___________i____I w = 3 lb down 2.5 0.5 0.2 2.2 0.6 F1 = ? up? Notice how we don't know in which direction the 2 support forces. We can guess which way they go and work out this problem on those assumptions. If we find a negative answer for a force it simply means we guessed wrong, and the force goes in the opposite direction. Make sure to clearly note what direction each force is assumed to be pointing. Lets assume in this problem that F1 and F2 both act upwards. The next step is to write an equation for the net force acting on the bar vertically (or "in the y direction"). Because we know that bar isn't moving, we can set this equation equal to 0: Fnet(y) = 0 = F1 + F2 - 7 - 3 Notice how all the forces which are pointing up are positive, and all those pointing down are negative. If there were forces acting in the horizontal on the bar we would do the same thing with those forces, but in this problem there are none so we can neglect that equation. This Fnet(y) equation has 2 variables, which means we have to find one more force equation in order to solve for these values. We can use the sum of the moments (or torques) around a point on the bar. While we can pick any point on the bar, the equation is easiest to solve if we pick the point where one of our unknown forces is acting. The moment of another force on the bar around that point is modeled by the equation: M = dFsin(x) Where F is the force, d is the distance from where the force acts to the point we picked, and x is the angle between the force and the horizontal axis. Notice that all the forces in this problem act normally to the horizontal, meaning sin(x) for all these forces is 1. This means we can ignore this part of the equation. Lets start by picking a point on the bar to sum moments around. We'll pick the point where F1 acts in this example. Now, we will calculate the moments applied by each force on the beam. If we start with F1 we see that d (the distance from F1 to F1) is 0, which eliminates that variable from our equation making it much easier to solve. We should end up with these moments: M(F1) = 0 M(w) = 1.5 lb*ft M(F2) = 0.7*F2 lb*ft M(L) = 20.3 lb*ft We now need to sum these together to find the net moment acting on the bar, just as we did with our forces. Before we do this though, we have to determine whether these moments are positive or negative. There are 3 easy steps for this: 1) Place your finger on the point where the force acts (i.e. if you want to find the sign of the moment caused by w, put your finger on the w). 2) Start drawing a circle with your finger with its center at the point where you are measuring the moment (in this case F1). You MUST start drawing the circle by following the direction the force you have your finger on. 3) Did your finger moving clockwise or counterclockwise? if you drew your circle clockwise, the moment that force creates is negative, if it was counterclockwise, the moment is positive. Our signed moments should be: M(w) = -1.5 lb*ft M(F2) = 0.7*F2 lb*ft M(L) = -20.3 lb*ft Now we can write our net moment equation: Mnet = 0 = -1.5 + 0.7*F2 - 20.3 Notice that this equation is also equal to 0 as the bar isn't moving. We can use this equation and the Fnet equation which we previously wrote to solve for F1 and F2. We should get: F1 = 31.1 lbs F2 = -21.1 lbs Because F2 is negative, we know that it doesn't act upwards as we first guessed, but downwards. Our final answer becomes: F1 = 31.14 lbs up F2 = 21.14 lbs down
Subject: Physics (Electricity and Magnetism)
Consider a +7x10^-6 C charge and a +3x10^-6 C charge at a 25cm radius. At what point in between these charges would a -5x10^-6 C charge have no net force. Assume all charges are in the same plane.
The key to every physics problem starts with drawing a diagram of the problem. Although I can't really draw anything here I'll do my best. Below is the situation as described in the problem. Each c represents a charge and they are demarcated as c1, c2, and c3, and the two distances that we need to solve for are labeled as well as r1 and r2. Laying out the known and unknown variables next to or on this diagram is important to ensure a good grasp of the question. (Also on exams teachers often give partial credit for a labeled diagram even if you solve the problem incorrectly so it's a useful habit to start problems this way.) c1------------------c2---------------------c3 c1 = +7x10^-6 C r1 = ? i____________i____________i c2 = -5x10^-6 C r2 = ? r1 r2 c3 = +3x10^-6 C i____________ ____________i 25cm Next we have to determine the directions of the forces that are acting on c2, the center charge. With a pair charges there are two options, they can either attract or repel each other. If the two charges are of the same sign, they will repel each other just like magnets that are the same pole. if they are opposite signs, they will attract each other. Lets consider the pair c1 and c2. Will they repel or attract? Because c1 is positive and c2 is negative these charges will attract each other. The same is true for the pair c2 and c3. This means there is a force pulling c2 towards both charges on either side. We should draw these forces that are acting on c2 into our diagram. We can call them F1 and F2. F1 F2 c1-----------<-------c2-------->-------------c3 c1 = +7x10^-6 C r1 = ? i____________i______________i c2 = -5x10^-6 C r2 = ? r1 r2 c3 = +3x10^-6 C i____________ ______________i 25cm Now, the question asks at what point -- or at what values of r1 and r2 -- does the net force on c2 equal 0. In order to determine this, we must first write an equation for the net force on c2. We do this simply by combining the F1 and F2. Make sure to properly represent the signs of the forces. If we assume that right is the positive direction (which we almost always do) then, looking at our diagram, we can see that F2 is pointing in the positive direction, and F! is pointing in the negative. Given this, our net force equation should be: Fnet = 0 = F2 - F1 We now have to use Coulomb's Law to replace F1 and F2 with their proper terms. Coulomb's Law is F = (k*q1*q2)/(d^2). k is a known constant, but for simplicity we can leave it as k here. F is the force between charges, q1 and q2 are the magnitudes of the charges, and d is the distance between them. Using this we can now replace F1 with (k*c1*c2)/(r1^2) and F2 with (k*c2*c3)/(r2^2). Our Fnet equation becomes: Fnet = 0 = (k*c2*c3)/(r2^2) - (k*c1*c2)/(r1^2) You may notice that we have 2 unknowns in this equation, r1 and r2. This means we need one more equation to find both values. Once again we can look to our diagram for help. We can see by the simple geometry of the diagram that: r1 + r2 = 25cm (it is important to keep track of what units you are working in) We can now use these two equations to solve for both r1 and r2. The solution we arrive at is that r1 = 15.1 cm and r2 = 9.9cm
Solve the following system of equations: 8x + 5y = 22, 15y - 3x = 12
Start by looking at the coefficients of each variable. x has the coefficients 8 and -3, and y has the coefficients 5 and 15. Then, decide if any of the x coefficients are multiples of each other. No? Okay, do the same for the y coefficients. Here we see something a little interesting. The second y coefficient is exactly 3 times the first. Knowing this makes solving this system much easier. The next step is to multiply every value in the first equation by 3, giving us: 24x + 15y = 66 Then we subtract the second equation from that. This might sound complicated at first but it's actually not too bad. All we need to do is combine the like terms in each equation. For example, start with the x terms we'll start with the 24x and then subtract the -3x from it, giving us 27x. Notice that 15y - 15y = 0 thus eliminating the y term from the resulting equation. We always want one variable eliminated after combining the equations. After subtracting you should get 27x = 54, and then we can easily see that x=2. Plugging this value back into either original equation allows us to easily solve that y=1.2 Example calculations: 3(8x + 5y = 20) -(-3x + 15y = 12) ----> x = 54/27 = 2 ----> -3(2) + 15y = 12 = -6 +15y ----> 15y = 18 ----> ______________ y = 18/15 = 1.2 27x + 0y = 54
needs and Eleanor will reply soon.