welcome YouTubers to another episode in my grammar hero Series in this video I'm going to work out 21 practice test questions that require you to perform calculations for the mechanical comprehension subtest of both the arm Services vocational aptitude battery that is the ASVAB as well as the pre-screening internet delivered computer adaptive test that is the pat on the actual test you'll have 20 minutes to answer 16 questions and in order to get the most from this video you'll want to pause the video after I read a question attempt to answer it on your own and then resume playing the video to check your solution as a quick reminder I want to point out that you're not permitted to use a calculator or reference sheet on the actual asab or PCAT including on the mechanical comprehension subtest that said uh the calculations that you're going to be performing on the mechanical comprehens and subtest are very simple and in my opinion by watching this video from start to finish you'll be exceptionally well prepared for this subtest for that reason I only recommend that you spend a day or two at most preparing for this subtest of the ASVAB and finally and perhaps most importantly I want to stress this again as I do in all my other videos do not spend more than $30 on test prep for the ASVAB truth be told everything you need to pass this test is freely available on my YouTube channel in fact I have a playlist for the mechanical comprehension subtest I'll put a link to this playlist in the description below if you go through it from start to finish that's all you really need to know for the mechanical comprehension subtest additionally I want to point this out if someone asks you to spend more than $30 on test prep including on tutoring for the ASB it's probably a scam and if that person has a foreign accent like an Indian accent it's also probably a scam guys this test is nothing but basic information and as I've said over and over again I've covered everything you need to know for this test freely on my YouTube channel so with all that being said let's go ahead and get started with these practice test questions this mechanical comprehension practice test question for the ASVAB and picat says what is the mechanical advantage of the pliers shown in the illustration so first of all here's our fulcrum and that's a clue that we're dealing with a simple lever here its mechanical advantage ma is going to be equal to the distance that the effort is from the fulcrum you can see that that is 6 in divided by the distance the load is from the fulkrum here's the load this is where we're going to grab onto something with the pliers we can see that that is 2 in away from the fulcrum so this is 6 / 2 which gives us a mechanical advantage of a three this mechanical comprehension practice test question for the ASVAB and Pat says an object moves on a frictionless plane at a velocity of 2 m a second if no external Force acts on the object what is the object's velocity after 10 seconds so this is testing your understanding of two concepts from physics the first of which is Newton's first law of motion that says an object will REM remain at rest unless a net force acts upon it likewise an object will remain in motion unless a net force acts on it well let's take a look at the velocity formula and see how this relates to Newton's first law of motion again velocity the final velocity V of T is going to be equal to your initial velocity V of o plus acceleration time time well let's look what we have in this formula we know our velocity our initial velocity V of O is 2 m a second there's no external force acting on this object so that means our acceleration in the velocity formula is zero and we know time is 10 seconds so let's plug those values in accordingly again 2 m a second plus again no acceleration in this case so that's going to be Zer for a Time 10 seconds for time so our final velocity V of T equals 2 0 * 10 anything time uh 0 is 0 so this is just our final velocity is going to be 2 m a second so this one is a again an object in motion will remain in motion unless a net force acts upon it such as acceleration deceleration or friction so this one is a this mechanical comprehension practice test question for the ASVAB and Pat says a car is moving at an initial velocity of 5 m a second and then accelerates uniformly at 10 m a second squared what is the car's veloc after 2 seconds so most of you can solve this one intuitively that said let's take a look at the velocity formula it says your final velocity V of T is going to be equal to your initial velocity V of o plus acceleration a * T which is time thankfully enough we were given all these things in this question uh all we have to do is go back and identify them its initial velocity V of O is 5 m a second uh it accelerates at 10 m a second squared so that's going to be a and for how long well we know T is 2 seconds so just plug those things in initial velocity was 5 m a second accelerated at 10 m a second squared for 2 seconds so this is simply going to be uh five 10 * 2 is 20 so our final velocity V of T is uh 20 + 5 are 25 M second so this one is going to be B of course this as fa Pat mechanical comprehension practice test question says two wheels as shown are connected by a belt F wheel a makes one complete rotation how many rotations will wheel B make so in this case we're not given the number of teeth on these uh Wheels instead we're given their diameters and we can quickly compute a gear r ratio using their diameters again this one has a diameter of 28 in this one has a diameter of 14 in we could reduce this ratio by a common factor of 14 uh 28 divid 14 is 2 and 14 divid 14 is 1 so they have a gear ratio of 2: one that means for every one ration this big gear makes this small gear is going to rotate two times times all right so this one is going to be C two rotations this asab Pat mechanical comprehension practice test question says if gear a makes 150 revolutions per minute how many RPMs does gear B make so you have to know this for this one uh the bigger the gear that is the bigger either its diameter is or the more number of teeth it has uh the slower it's going to rotate than a gear that has fewer teeth or a smaller diameter so by looking at this we can see gear a has 36 teeth and gear B has 18 teeth therefore we can say gear a is going to rotate uh more slowly than gear B now let's go ahead and calculate that we could do that using a quick proportion here again gear a has 36 teeth uh gear B has 18 Teeth Let's go ahead and divide these by a common factor of 18 to get our ratio 36 ID 18 is 2 18 ID 18 is 1 so the ratio and the number of teeth on these gears is 2 to one what does that mean that means for every one rotation this gear makes uh this smaller gear is going to make two rotations in other words if gear a is spinning at 150 RPMs uh gear B is going to be spinning twice as fast 150 x 2 is 300 that is at 300 RPMs so this one is going to be d uh gear B is going to be rotating or spinning at 300 RPMs again we calculated that it spins or rotates twice for every one time gear a rotates so that is that one pretty simple this mechanical comprehension practice test question for the ASAP and Pat says if gear a is rotated by a motor at a rate of 100 RPMs how many revolutions will gear C make per minute so in order to figure this one out we have to calculate the gear ratio between gear a and gear B and then finally between gear B and gear C since all these gears are inter meshed uh generally speaking the more teeth a gear has the more slowly it spins compared to a gear with fewer teeth so let's start with gear a and gear B we know gear a spins at 100 RPMs to find its ratio we're simply going to do 20 over 60 uh this reduces to 1/3 what does this mean for every three revolutions uh gear a makes gear B is going to make one revolution so if gear a is uh making 100 revolutions per minute gear B is going to do 100 / 3 revolutions per minute all right so now that we know uh how fast gear B is rotating let's do the same thing over here again gear B has 60 teeth whereas gear C has 10 teeth uh this reduces to 6 over one what does this mean for every one revolution gear B makes gear C is going to make six revolutions so let's figure that out doing the math here we're going to do 100 over3 * 6 over 1 uh we could cross reduce three goes into three one time three goes into six two times this is simply 100 over one which is just 100 * 2 over one which is just two 100 * 2 is 200 so gear C is rotating at approximately 200 RPMs D this mechanical comprehension practice test question for the as Fab Pat says this the driver gear Gear a turns clockwise at a rate of 60 RPMs in what direction does gear B turn and at what rotational speed so first of all inner mesh gears always rotate in opposite directions so if gear a is rotating clockwise that means gear B is going to be rotating counterclockwise likewise you have to know that the more teeth a gear has the more slowly it's going to rotate compared with a gear that has has fewer teeth uh gear B has more teeth than gear a so it's going to rotate more slowly than gear a if gear a is rotating at 60 RPMs gear B is going to be rotating more slowly than that so all these answer choices with 120 in them are incorrect let's calculate our gear ratio our gear ratio is the number of teeth on the driven gear uh again gear B is being driven so that's 40 over the number of teeth on the driving gear again the drive diving gear is gear a that has 20 cross these out this reduces to a ratio of 2: one what does this mean every two times gear a rotates gear B is going to rotate one time how are we going to calculate that we're simply going to do 60 over 2 which is 30 so gear B is rotating at 30 RPM so this one is D counterclockwise and at 30 RPMs this mechanical comprehension practice test question for the ASAP and PCAT says this the three pulleys are connected by Rubber belts the two pulleys at the top have the same diameter while the pulley below is twice their diameter if the driver pulley at the upper left is turning clockwise at 60 RPMs at what speed and in what direction is the large bottom pulley turning so here's our Drive pulley right here let me say this uh the size of the pulley corresponds to how fast it spins the bigger the pulley the more slowly it's going to spin compared to a pulley with a smaller diameter uh so in other words if this pulley is spinning at 60 RPMs this one has the same diameter so it's also going to spin at 60 RPMs that said the diameter of this bottom pulley is twice the diameter of this pulley here so it's going to spin half as fast as this pulley if this one's spinning at 60 RPMs this one's going to spin at 30 RPMs so all these anal choices with 120 in them are incorrect now we just have to figure out what direction this bottom pulley turns in again this one's going clockwise so this belt is moving this way and you can see that's going to come around this way and spin this one this way which is counterclockwise and that's going to continue to spend this pulley in uh the direction of B which is counterclockwise so this one is going to be B it's spinning at 30 RPMs and it's also spinning in Direction B which is counterclockwise this mechanical comprehension practice test question for the ASVAB and Pat says how much force is needed to apply 50 foot pounds of torque to a bolt using a two foot long wrench so there is a Formula you can use to quickly solve this one notably t torque is equal to R * F where R refers to the length of the lever arm and F refers to force let's look through this problem and see what information we have we know that we want to apply 50 foot-pounds of torque so T is going to be 50 uh our lever arm is the length of our wrench uh and we know that is 2 feet so R is two let's plug these values in and then as you'll see we'll solve for Force very quickly again 50 ft-lbs of torque equals uh 2 foot long lever arm time F if we divide both sides by two we'll get our answer that is we'll know how much force uh is needed 50 divid by 2 is 25 so you're going to need d25 lbs of force in this case this mechanical comprehension practice test question for the asab and pet says if 100 pounds of force are applied over an area of 2 square in pressure would be so you could solve this one using common sense that said there is a formula for this one as well notably pressure is equal to force divided by area in this case the force is 100 PB and the area is 2 square in 100 divid 2 is 50 so this one is going to be 50 psi which is answer Choice B of course this mechanical comprehension practice test question for the ASVAB and pet says if a force of 200 lb is exerted over an area of 10 square in what is the pressure in terms of PSI that is pounds per square inch so for this one you can solve it using common sense that said there is a Formula that you can use as well that is pressure equals force / area in this case we're solving for p the force according to the problem is 200 lb over an area of 10 square in uh this is very easy to reduce we cross out these corresponding zeros this becomes 20 over 1 20 / 1 is just 20 so in other words uh the pressure is going to be C20 PSI this mechanical compreh and practice test question for the ASAP and piket says the surface area piston a is 4 Square in and the surface area piston B is 16 square in piston a has pushed down 12 in how far will piston B move up so according to the problem the surface area of piston a is Four Square in and the surface area of piston B is 16 square in in other words piston B has four times the surface area of piston a in addition we know that we're pushing down piston a 12 in and we want to know how far up piston me B moves in response well as it happens we can use some uh fluid mechanics to solve this one it's very simple Pascal's law states that uh pressure is transmitted equally in all directions in a fluid which means that if piston B has four times the surface area of piston a that means it's going to move up 1/4 the amount that piston a is pushed down since it has four times the surface area so we're going to do 1/4 * 12 if you haven't done this math in a while we can multip multiply 1/4 by 12 by placing 12 over 1 again 12 / 1 is 12 then when you multiply two fractions you just multiply straight across 1 * 12 is 12 4 * 1 is 4 12 / 4 is 3 so it's going to move Up 3 in in response so this one is B this mechanical comprehension practice test question for the ASAP and PCAT says this the surface area of the small piston is 4 Square in while the surface area of the large piston is 8 square in if the small piston is pushed down 8 in how far will the large piston move up so again the surface area of this small piston is 4 Square in the surface area of this large piston is 8 square in as you can imagine uh the distance that each piston moves corresponds to the ratio of their surface areas and again this one is being driven down 8 in uh so let's go ahead and calculate the ratio of the surface areas of these Pistons again this large piston has the surface area of 8 square in the small piston has a surface area of Four Square in cross out these units this gives us a ratio of 2 to one in their surface areas in other words for every two in that this piston moves down or up this piston right here is going to move 1 in correspondingly so if this one moved 8 in down this one is going to move 8 / two are 4 Ines up so this one is going to be C this mechanical comprehension practice test question for the ASVAB and PCAT says this what is the amount of force needed to keep the barrel from moving if we look at our diagram you can see that we have 150b Barrel on an inclin plane that is 9 ft long and goes up 3 ft vertically now for this subtest of the ASB you really don't need to know any formulas all these questions can be solved using nothing more than Common Sense uh that said uh the formula does make sense so let's take a quick look at it the force that we need to keep this Barrel from moving is going to be equal to the weight of the barrel times s Theta that is the angle of incline right here right so what is the weight of the barrel according to the diagram it's 150 lb what is sin Theta well let's think about that for a second and you do have to know some basic trig to answer this one again I remember s cosine and tangent using the pneumonic device sooa again s so s is equal to opposite over hypotenuse what is the opposite of theta here it's 3T what is the hypotenuse of this right triangle it's 9t so s is 3 over 9 which we know reduces to 1/3 so this becomes 150 * 1/3 write 150 as a fraction by placing it over one multiply these two fractions now this becomes uh 150 * 1 which is 150 over 1 * 3 150 divided by 3 is simply 50 uh so this one is going to be a it's going to take 50 lbs of force to keep this Barrel from sliding down this incline plane this mechanical comprehension practice test question for the ASAP and Pat says this what is the amount of force needed to keep the barrel from moving we're going to use a very simple formula to calculate this one if you didn't know the formula you could also calculate this one using Common Sense we're going to say the force needed to keep the barrel from moving it's going to be equal to the weight of the barrel times sign of theta where Theta is this angle of incline in this incline plane well what is the weight of this Barrel it is 100 pounds you can see that right here how do we find S Theta well uh s Theta is easy to calculate if you remember the pneumonic device SOA TOA again so says s is opposite over hypotenuse the opposite of theta here is four hypotenuse of this right triangle is 20 and this reduces to 1 15 so sin Theta is simply 1 again you could write 100 as a fraction by placing it over one and you just multiply straight across this becomes 100 over 5 100 ID 5 is 20 in other words it's going to take 20 pounds of force to keep that Barrel from rolling down the inclin plane mechanical comprehension practice test question for the asab and PCAT says this a cable is pulling a 320 lb box up an incline that's 16 feet long if 80 pounds of force is used to move the box up the incline how tall is the incline so to answer this one you have to know the formula force is going to be equal to the weight of the object times s Theta or sign data refers to the angle of incline how much force is needed well according to the problem they needed 80 lb of force how much does the object weigh 320 lb uh what is s Theta well s cosine and tangent can be remembered using the pneumonic device SOA TOA again so refers to S so sin Theta is opposite over hypotenuse what is the opposite of theta here it's x what is the hypotenuse of this this right triangle it's 16 so sin Theta is X over 16 to make this math a little bit easier we can cross reduce 16 goes into 16 one time uh 16 goes into 320 how many times Well 16 goes into 32 uh two times just bring up that zero so that reduces to 20 this becomes 80 equals what is 20 * X over 1 X over 1 x / 1 is just X so this is just 20x to get X by itself we're going to divide both sides of this equation by 20 this crosses out leaving you with X on this side 80 / 20 is 4 so X refers to the height of this inclined plane we now know X is four so this one is going to be C4 feet this as FB Pat mechanical comprehension practice test question says in this diagram of a simple gearbox the gear ratios for each pair of Gears is given what will be the output speed if the input speed is 240 RPM so our input speeds right here it's 240 uh don't pay attention to the size of the gears in this gear box instead we're going to be calculating the output speed in terms of RPMs and we're going to use these gear ratios to figure out the output speed so let's get started we know our input speed is 240 and this gear and this gear have a ratio of 4:1 to figure out how many RPMs this one is spinning at we're going to divide this by four which which is 60 so this one's spinning at 60 RPMs again this gear and this gear have a ratio of 3 to one we know this one's spinning at 60 to figure out how fast this one's spinning we're going to divide it by three 60 divided by 3 is 20 so we know this one is spinning at 20 RPMs and finally uh this gear and this gear have a ratio of 2 to one this one's spinning at 20 to figure out how fast our output speed is we're going to do 20/ by two which is 10 so our output speed is 10 RPMs which is answer Choice a this mechanical comprehension practice test question for the as Fab and Pat says this in case a both blocks are fixed in case b the load is hung from a movable block ignoring friction what is the required force to move the load in both cases so let's look at case b we have a mechanical advantage of 2: one here because we have two tension points we have a tension Point here and here so 500 Newtons is going to be required there and 500 Newtons is going to be required there in total Force two is only going to require 500 Newtons of force so all these answer choices that have F2 being a th000 are incorrect cross them out now case a gives us no mechanical advantage so you can see takes a th000 Newtons here it's going to take a th000 Newtons here and it's going to take a th000 Newtons of force here so this one is going to be C Force One is going to be a th000 whereas Force two has a mechanical advantage of uh two to one so it's going to take 500 Newtons of force so again C this mechanical comprehension practice test question for the As faab and Pat says if an effort of 2 lbs is applied on one side of the fulcrum to lift a load of 8 lb on the other side of the fulcrum what is the mechanical advantage of the lever so for the asab and pet you will have to calculate mechanical advantage thankfully enough that's pretty simple to do I put the formula down here for you it's equal to output force over input force which is the same thing as your load divided by your effort in this case we're told that we have a load of 8 pounds and it takes an effort of two pounds uh to lift the load so this is 8 divided by two which is four in other words uh this lever gives us a mechanical advantage of four so the answer to this one is B this challenge question for the mechanical comprehension subtest of the ASVAB and Pat says how much force is needed to balance the system so before we get started I want to point a few things out uh this triangle right here is What's called the fulcrum and as you can see we have two weights sitting on the right hand side of our system and we're going to be applying our force over here now there is a Formula you need to know to solve this one and that's right here it's very simple don't get overwhelmed by it it says the force applied are the weight of the load times the distance from the center of the load to the fulcrum is equal to the force applied by the effort times the distance from the center of the effort to the fulcrum well what does that mean and in this case we're talking about weights let's start with this 5 kg weight according to this part right here we take our weight and we multiply it by the distance from the center of the load to the fulcrum well we can see that here's the center of the fulcrum this 5 kg weight sits directly above the fulcrum so its distance away from that is zero plus let's take a look at this 20 kg weight now what are we going to multiply this by the distance from the center of the load to the fulcrum here's the center of our load here's our fulcrum we can see that that is a distance of 2 m so we're going to multiply this by two all right so we just filled in the left hand side of our equation now let's move on to the right hand side over here it's going to be equal to the force applied by the effort this is the force we're applying over here so that's just just going to be the letter variable F times the distance from the center of the effort to the fulcrum you can see that that is right here our force is 5 m away from the fulcrum so we're going to multiply this by five all right so all we have to do is solve this equation for f and we'll have our answer uh f * 5 is 5 F 5 * 0 is 0 now why does this go to zero well if you take a look this 5 kg weight sits on both sides of the fulcrum so it has no impact on the balance of the system all right 20 * 2 is 40 so we're going to be solving for f the next thing we're going to do is divide both sides by five this crosses out leaving you with f over here 40 divided by five is 8 so our force is going to be 8 kgs what does that mean that means we have to apply 8 kgs of force over here to balance this system so this one is A8 kgs Guys these questions are very easy to solve as long as you know that you multiply the force or the weight times the distance it is from the center of the fulcrum and set them equal to each other and you'll get your answer in less than two minutes so make sure you can solve these these do frequently appear in the mechanical compress comprehension subtest of the asab and Pat this mechanical comprehension practice test question for the ASVAB and Pat says how much force is needed to balance this system of course I included the formulas you need to know to solve this one notably that the force applied are the weight of the load times the distance from the center of the load to the fulcrum is equal to the force applied by the effort times the distance from the center of the effort to the fulcrum in other words the force or the weight of the load times the distance it is uh to the center of the fulcrum is going to be equal to the force of the effort times the distance that effort is from the center of the fulcrum this is very easy to fill in don't get overwhelmed by this again we're going to be applying our Force right here let's start with our weight our weight is 10 kg so that's going to be 10 how far away is that from the center of this fulcrum here it's 1 M so this is 10 * 1 equals force of effort well we're going to be solving for that how much force is needed we have to solve for that so we're going to leave that as F of e times its distance from the center of this fulcrum well we can see that that is 2 m so this is f of e * 2 again we're solving for f of e so 10 * 1 is 10 = 2 F of e to get F of e by itself we're going to divide both sides by two this this crosses out leaving you with f of E = 10 / 2 or 5 so this one is C5 kgs it's going to take five kgs of force uh to balance this system