The Law Posted October 8, 2009 Report Share Posted October 8, 2009 I'm sure i have a great proportion of the threads in the MCAT forum. You know what they say about third times though... right? Anyway. Does a math person want to explain to me what the what a 30-60-90 triangle is? I doubt I'm going to encounter it on the MCAT, but I guess I should know - just in case. Link to comment Share on other sites More sharing options...
eng_dude786 Posted October 8, 2009 Report Share Posted October 8, 2009 Hi Law, The 30-60-90 triangles are special triangles with angles of 30, 60 and 90. What they allow you to do is calculate Sine and Cosine of 30, 60 and 90 easily by using simple trigonometry relations. Only specific triangle 'length' sides give a triangle of 30, 60 and 90. So essentially the goal is that you can remember the side lengths and the angles of the triangle vertices you should easily be able to calculate the Cosine and Sine (and even Tan) of common angles such as 30, 60 and 90. An equivalent triangle is 45-45-90. And that allows you to easily calculate the Cosine and Sine of angles of 45, 45 and 90. However the special triangles only give you answer in fractional form. So answers like root(3)/2 or root(2) So in summary, the only need for special triangles is to calculate cosine and sine of specific common angles. However for the MCAT, I recommend you just memorize to 2 decimal places the specific angles. Like sine of 60 is 0.87 No need to worry about special triangles as long as you got the common angles memorized. Link to comment Share on other sites More sharing options...
dan0105 Posted October 8, 2009 Report Share Posted October 8, 2009 what! memorizing the two decimal places. Screw that. learn the triangles. They make way more sense once you get them down, are easier to retain in memory, and most of physical chem questions have things that cancel out easily if you wait until the end to plug in the numbers and solve them. Link to comment Share on other sites More sharing options...
The Law Posted October 8, 2009 Author Report Share Posted October 8, 2009 ENG DUDE... you're back!! How are you!! Link to comment Share on other sites More sharing options...
The Law Posted October 8, 2009 Author Report Share Posted October 8, 2009 Okay so what exactly are these special dimensions. The question I got was: A man entered a cave and walked 100 m N. He then made a sharp turn 150 degrees to the West and walked 87m straight ahead. How far is the man from when he entered the cave? (Note: Sin 30 = 0.5, cos 30 = 0.87). Link to comment Share on other sites More sharing options...
AG22 Posted October 9, 2009 Report Share Posted October 9, 2009 Okay so what exactly are these special dimensions. The question I got was: A man entered a cave and walked 100 m N. He then made a sharp turn 150 degrees to the West and walked 87m straight ahead. How far is the man from when he entered the cave? (Note: Sin 30 = 0.5, cos 30 = 0.87). Okay, so. THe man walks 100 m straight ahead. He then turns 150 degrees to the west. This means, he turns 90 deg to the left, then 60 degrees further to the left and walks 87 m. So he has essentially now made a triangle, with a 30degree angle between the 100 m north and 87 m westward. So now, the q is asking, how much to the left of the starting point (cave's entrance) is this man at the end of his 87 m walk? (let's call this x) sin of 30 = opp/hyp = 0.5 thus, 0.5 = x/87m thus x = 87/2 so if we assume 87 is approx 90, the man has travelled approx 45 m to the left of the entrance. hope that helps! OR, i could be totally wrong with my interpretation of the q and someone else may be kind enough to step up to the plate oh. p.s.! i actually did end up memorizing. But I didn't really need it on the real deal. I knew what sin and cos of 30, 60 and 90 degrees were, in decimals and fractions. Link to comment Share on other sites More sharing options...
The Ace of Spades Posted October 9, 2009 Report Share Posted October 9, 2009 So in summary, the only need for special triangles is to calculate cosine and sine of specific common angles. However for the MCAT, I recommend you just memorize to 2 decimal places the specific angles. Like sine of 60 is 0.87 No need to worry about special triangles as long as you got the common angles memorized. Don't bother memorizing to two decimal places. Whenever trigonometry is used on the MCAT, they provide you with numerical values of those sine and cosine values. Link to comment Share on other sites More sharing options...
The Law Posted October 11, 2009 Author Report Share Posted October 11, 2009 It's pretty easy to memorize if you wanted to! But yeah, usually they give you the value. Link to comment Share on other sites More sharing options...
terryann Posted October 11, 2009 Report Share Posted October 11, 2009 Go Law Go... Link to comment Share on other sites More sharing options...
The Law Posted October 11, 2009 Author Report Share Posted October 11, 2009 Thanks terryann! Does anyone want to try to explain what friction is to me? I just got a question wrong so I don't think I'm really understanding this concept well. In examkrackers: If the rear wheels of the truck pictured below drive the truck forward, then the frictional force on the rear tires due to the road is (pretend the O below is the rear tire): A<--O-->B A) Kinetic and in the direction of A Kinetic and in the direction of B C) Static and in the direction of A D) Static and in the direction of B Answer=D I understand now that it's static friction because if you consider the point where the tires contact the road, that point is not moving relative to the road. I don't really understand how to determine which direction the frictional force acts here. I know friction opposes relative motion... I'm guessing I just have to remember car tires, the frictional force is the force making the car go forward. ok draw yourself a line and a circle tangent to it like so O _____ but obviously touching. then draw arrows around the circle showing the clockwise turning of the tire. at the bottom of the circle where it's touching the line, the arrow points to the left (like so <--- ) OOO, that makes sense!! Thanks eshgh. Link to comment Share on other sites More sharing options...
The Law Posted October 11, 2009 Author Report Share Posted October 11, 2009 Another question, about Hooke's law this time: In many harbors, old automobile tires are hung along the side of wooden docks to cushion them from the impact of docking boats. The tires deform in accordance with Hooke's law. As a boat is brought to a stop by gently collding with the tires, the rate of deceleration of the boat (until the boat stops): A) is constant decreases C) increases D) inreases, then decreases Answer= C So, my thinking is: Boat is causing force on the tires. This force is directly proportional to the acceleration of tire stretching (and also, deceleration of the boat). So as the boat presses against the tire, the deceleration of the boat increases (since the force increases the more the tires are squished away from their original shape - F = k x). However, I picked D because I thought that after a point, the tires will then reform their shape and so the deceleration will go down. Clearly the last line of thinking is wrong since that's not the answer... but can someone help me find my error in thinking! Merci! Link to comment Share on other sites More sharing options...
AG22 Posted October 11, 2009 Report Share Posted October 11, 2009 Another question, about Hooke's law this time: In many harbors, old automobile tires are hung along the side of wooden docks to cushion them from the impact of docking boats. The tires deform in accordance with Hooke's law. As a boat is brought to a stop by gently collding with the tires, the rate of deceleration of the boat (until the boat stops): A) is constant decreases C) increases D) inreases, then decreases Answer= C So, my thinking is: Boat is causing force on the tires. This force is directly proportional to the acceleration of tire stretching (and also, deceleration of the boat). So as the boat presses against the tire, the deceleration of the boat increases (since the force increases the more the tires are squished away from their original shape - F = k x). However, I picked D because I thought that after a point, the tires will then reform their shape and so the deceleration will go down. Clearly the last line of thinking is wrong since that's not the answer... but can someone help me find my error in thinking! Merci! Okay, so they want to know until the boat stops right? you would be correct had they not said until the boat stops. B/c in that case, the tires would reform their shape and push the boat backwards away from the dock, so the acceleration would decrease then increase (increase once the tires sprung back up). But by the time the boat docks, they immediately tie it up or anchor it or w/e so there is some motion backwards but not a whole lot. The wording of the question is very important! I learned that the hard way hehe. So the best answer, is C, in that until the boat stops, the acceleration of the boat (ie deceleration) increases. Link to comment Share on other sites More sharing options...
AG22 Posted October 11, 2009 Report Share Posted October 11, 2009 Thanks terryann! Does anyone want to try to explain what friction is to me? I just got a question wrong so I don't think I'm really understanding this concept well. In examkrackers: If the rear wheels of the truck pictured below drive the truck forward, then the frictional force on the rear tires due to the road is (pretend the O below is the rear tire): A<--O-->B A) Kinetic and in the direction of A Kinetic and in the direction of B C) Static and in the direction of A D) Static and in the direction of B Answer=D I understand now that it's static friction because if you consider the point where the tires contact the road, that point is not moving relative to the road. I don't really understand how to determine which direction the frictional force acts here. I know friction opposes relative motion... I'm guessing I just have to remember car tires, the frictional force is the force making the car go forward. It's static b/c of what you've said. And if you think about it, there isn't anything really touching the car and moving it forward except the road is there? In that case then, friction pushes the car forward. B/c the way I think of it is like this, frictional force opposes an applied force. When there is no force applied, the frictional force is the only force that is doing all the work and changing the direction. One may say what about the fact that the car is accelerating?? Isn't F = mcar times acceleration? So isn't there technically a force on the car?? But in this case, F = net force, which is equal to Friction road. Fnet = Ffriction = ma. Because there is no other external force on the system (i.e. car). Thus, force of friction is what causes the car to move in the direction it is shown in the picture. Even if the direction was reveresed, so instead of Right, the car was going left, the force of friction is what causes the motion. I don't know if i've conveyed my thoughts clearly enough. Perhaps someone else with better teaching skills will step up to the plate? Link to comment Share on other sites More sharing options...
The Law Posted October 11, 2009 Author Report Share Posted October 11, 2009 Okay, so they want to know until the boat stops right? you would be correct had they not said until the boat stops. B/c in that case, the tires would reform their shape and push the boat backwards away from the dock, so the acceleration would decrease then increase (increase once the tires sprung back up). But by the time the boat docks, they immediately tie it up or anchor it or w/e so there is some motion backwards but not a whole lot. The wording of the question is very important! I learned that the hard way hehe. So the best answer, is C, in that until the boat stops, the acceleration of the boat (ie deceleration) increases. Yeah, thinking about it I started confusing myself because then I started thinking... "Wait, if it springs out, the boat will start accelerating away from the dock! How is it going to stop?"... I tend to make things more complicated than they have to be. That's one of my biggest problems, too reductionist... blame my biochem background. It's static b/c of what you've said. And if you think about it, there isn't anything really touching the car and moving it forward except the road is there? In that case then, friction pushes the car forward. B/c the way I think of it is like this, frictional force opposes an applied force. When there is no force applied, the frictional force is the only force that is doing all the work and changing the direction. One may say what about the fact that the car is accelerating?? Isn't F = mcar times acceleration? So isn't there technically a force on the car?? But in this case, F = net force, which is equal to Friction road. Fnet = Ffriction = ma. Because there is no other external force on the system (i.e. car). Thus, force of friction is what causes the car to move in the direction it is shown in the picture. Even if the direction was reveresed, so instead of Right, the car was going left, the force of friction is what causes the motion. I don't know if i've conveyed my thoughts clearly enough. Perhaps someone else with better teaching skills will step up to the plate? You conveyed them well enough. That's a good way to think about it! The frictional force is the net force moving the car, nothing else is acting on it... so I need to think of it in terms of what's going to happen to the car, and that will let me know. Thanks for your help! Link to comment Share on other sites More sharing options...
avenir001 Posted October 11, 2009 Report Share Posted October 11, 2009 You conveyed them well enough. That's a good way to think about it! The frictional force is the net force moving the car, nothing else is acting on it... so I need to think of it in terms of what's going to happen to the car, and that will let me know. Thanks for your help! also if u for eg look at the right rear tire from the side and imagine it turning (in a clockwise direction over the road surface), at the point where the tire touches the road surface, the clockwise turning motion translates into a leftward tangential motion over the road surface..and since the force of friction opposes this leftward motion, the force of friction must be in the opposite direction (to the right towards . Link to comment Share on other sites More sharing options...
The Law Posted October 11, 2009 Author Report Share Posted October 11, 2009 "leftward tangential motion" ... confusion... Link to comment Share on other sites More sharing options...
avenir001 Posted October 11, 2009 Report Share Posted October 11, 2009 why doesn't that surprise me well the road surface is tangent to the tire right? as in...a line being tangent to a circle? tangential motion is just motion at the point of contact where u can imagine the tire sliding to the left along the road surface...with the friction opposing that motion. Link to comment Share on other sites More sharing options...
The Law Posted October 11, 2009 Author Report Share Posted October 11, 2009 why doesn't that surprise me well the road surface is tangent to the tire right? as in...a line being tangent to a circle? tangential motion is just motion at the point of contact where u can imagine the tire sliding to the left along the road surface...with the friction opposing that motion. Okay, now I can picture the "tangential motion"... but what I don't understand is why this is leftward? Link to comment Share on other sites More sharing options...
avenir001 Posted October 11, 2009 Report Share Posted October 11, 2009 ok draw yourself a line and a circle tangent to it like so O _____ but obviously touching. then draw arrows around the circle showing the clockwise turning of the tire. at the bottom of the circle where it's touching the line, the arrow points to the left (like so <--- ) Link to comment Share on other sites More sharing options...
The Law Posted October 11, 2009 Author Report Share Posted October 11, 2009 How about this passage: Statistically speaking, traveling on US highways is more dangerous than airplane travel. At the high speeds achieved by vehicles on the highway, turns must be very gradual. As a safety precaution, highway turns are banked toward the inside. Fed guidelines speccificy highway curve speed limits based upon the angle of the bank, coefficient of friction between vehicle and the pavemeent, and the radius of curvature of the turn. The radius of curvature of a circle is the radius of a circle that would be circumscribed by the vehicle if it were to complete a full circle. (LOL - my attempt to recreate the image) The diagaram above shows a vehcile on a highway curve moving ina direction out of the page and turning to the driver's left. Which of the following would require the bank angles which curently exist on highway to be increased? A) The average mass of vehicle increases ^mass decreases C) Curve speed limits are increased D) Curve speed limits are decreased Answer: D Link to comment Share on other sites More sharing options...
avenir001 Posted October 11, 2009 Report Share Posted October 11, 2009 i have no idea why it's D...i'd choose C cuz if your angle is higher, u're gonna need a higher critical speed to keep going in your circular path Link to comment Share on other sites More sharing options...
The Law Posted October 11, 2009 Author Report Share Posted October 11, 2009 LOL, oops the answer is C... not D! My bad. "The mass of the vehicles will not affect the required bank angle because centripetal force, frictional force, and the force down the incline due to gravity are all proportional to mass. A greater speed limit on the curves would require a greaterbank angle in order to keep the cars from sliding off the road." That makes sense, I mean I would pick C from intuition... but I don't really understand the details about why. Link to comment Share on other sites More sharing options...
avenir001 Posted October 11, 2009 Report Share Posted October 11, 2009 *eyeroll* well in that diagram, B is the centripetal force and is equal to N sin(theta) where N is the normal force perpendicular to the surface. so it's like Fc = N sin(theta) = mv^2/r. For a given angle, v is the speed you need to continue in your circular path. if your speed is higher, u'll drift up the incline and if it's lower, u'll slip down. with a higher speed, u need more of an incline to keep going in your circular path. with a very low speed, u don't even need much of an incline to be safe. Link to comment Share on other sites More sharing options...
AndreaM Posted October 12, 2009 Report Share Posted October 12, 2009 Okay, so. THe man walks 100 m straight ahead. He then turns 150 degrees to the west. This means, he turns 90 deg to the left, then 60 degrees further to the left and walks 87 m. So he has essentially now made a triangle, with a 30degree angle between the 100 m north and 87 m westward. So now, the q is asking, how much to the left of the starting point (cave's entrance) is this man at the end of his 87 m walk? (let's call this x) sin of 30 = opp/hyp = 0.5 thus, 0.5 = x/87m thus x = 87/2 so if we assume 87 is approx 90, the man has travelled approx 45 m to the left of the entrance. hope that helps! OR, i could be totally wrong with my interpretation of the q and someone else may be kind enough to step up to the plate oh. p.s.! i actually did end up memorizing. But I didn't really need it on the real deal. I knew what sin and cos of 30, 60 and 90 degrees were, in decimals and fractions. You got the concept right, but the answer wrong. The 100m side is the hypotenuse and the 87 is the adjacent. sin 30 = opp/hyp = 0.5 --> x/100=0.5 --> x=50 Link to comment Share on other sites More sharing options...
AG22 Posted October 12, 2009 Report Share Posted October 12, 2009 You got the concept right, but the answer wrong. The 100m side is the hypotenuse and the 87 is the adjacent. sin 30 = opp/hyp = 0.5 --> x/100=0.5 --> x=50 Ah! Crap! You're right. 87 m can never be the hypotenuse. longest side = hypotenuse. Thanks!! Link to comment Share on other sites More sharing options...
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