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in this video were going to focus on solve intention problems so in this question a rope is used to lift a 50 kilogram box with an upward acceleration of 2.3 m/s squared what is the tension in a rope so lets say this is the box and the rope is lifting up the box the force that acts along a rope is the tension force so how can we calculate this tension force what can we do well we need to set up a free body diagram we need to identify all of the forces that are in play so we have the upward tension force the downward weight force and also the net force now the net force is in the upward direction because the whole object is being lifted up so the net force in the Y direction which can also be written this way its going to be the tension minus the weight force so its t minus W our goal is to find T now the net force is equal to MA since the forces in the Y direction the acceleration is in the Y direction and so we got to add W to the other side so ma y plus W is equal to the tension force and the weight force is simply mg so if we take out M the tension force is going to be the mass times G Plus the vertical acceleration so this is the equation that we need so now lets get the answer for Part A so the tension force is going to be equal to the mass of 50 kilograms times the gravitational acceleration plus the upward acceleration of two point three meters per second squared so two point three plus nine point eight thats going to be nine point eight plus two is, 11 point eight and then if you add point three to that thats going to be, 12 point one and then we got to multiply that by 50 so the tension force is 605 Newtons now if you want to compare it to the weight force the weight force is mg so thats gonna be 50 times 9.8 which is 490 so notice that when the rope is used to lift up the object with an upward acceleration the tension force is greater than the weight force but now what if the Rope is being used to allow the box to slowly descent intuitively we know that the tension force should be less than a way force if the box is descending with a a downward acceleration so lets get the answer for Part B so M is 50 G is not 20 but the acceleration because its downward its going to be negative 0.75 instead of positive 4 in 75 so 9.8 minus 0.75 thats nine point zero five and then times 50 this is going to be 450 two point five Newtons so as you can see the tension force is less than the weight force when the object is slowly descending with a downward acceleration now lets work on
this problem what is the tension in the two ropes in the picture shown below now notice that the crate or the box whatever you want to call it its in equilibrium its at rest so therefore the sum of all forces in the and in the Y direction must add to zero so when you see a problem like this you want to break down t1 and t2 and into its components t2 has an X component and a Y component t1 also has an X component and a Y component t1 Y and the crate also has a weight force now because the object is at rest because its in equilibrium the net force in the X direction and in the Y direction must be zero so lets focus on the forces in the X direction this one is in the positive x direction so thats positive t2x this one is in a negative x direction so its negative t1x and because the object is at rest the net force in the X Direction is zero so if we add T 1 X to both sides we can see that T 1 X is equal to t2 X now t ax is T cosine beta T y is T sine theta so T 1 X is gonna be t1 cosine theta T 2 X is t2 cosine theta you can call this cosine theta2 if you want in cosine theta one to distinguish these two angles so for T 1 the angle thats associated with it is 60 so t1 cosine 60 is equal to t2 cosine 30 now cosine 60 in degree mode thats 0.866 action – thats 1/2 thats 0.5 cosine 30 is 0.866 so lets divide both sides by 0.5 0.866 divided by 0.5 thats 1.73 – so t1 is 1.73 – times t2 so lets save this equation for later now lets focus on the forces in the y-direction so we have two upward forces t1 Y and T 2 I so theyre both going to be positive and we have a weight force in the negative y-direction so thats going to be negative W now because the object is at rest and that force in the y-direction is 0 so Im going to add W to both sides so W is equal to t1 y + t2 Y so these two upward tension forces must balance or support the download weight force so the weight which is basically mg and thats equal to t1 Y which is t1 sine theta 1 plus t2 y which is t2 sine theta 2 so M the mass is 60 and G is 9.8 and t1 we dont know what that is right now but theta1 is 60 and theta2 is 30 now sixty times 9.8 thats 588 and sine 60 is 0.866 sine 30 is 0.5 so now what Im going to do is Im going to replace T 1 with 1.73 2 T 2 so we got to solve this using substitution anytime you have two variables you need two equations to find or
to solve those two variables so now this is gonna be 588 which is equal to twenty eight six six times one point seven three two if you multiply those two numbers youre gonna get a number thats if rounded one point five and then lets add this to it so 1.5 plus 0.5 thats – so 588 is equal to 2 times T 2 so 588 divided by 2 is 294 so thats the tension force in this rope now using this equation we can find C 1 so T 1 is going to be one point seven three two times 294 so just take this number plug it into this equation and so T one is about five hundred and nine point two Newtons so now we have the two tension forces so this is more these are the answers but now lets check our work lets make sense of everything so heres a 60 kilogram box and it has a download weight force of 60 times 9.8 which is as we know 588 Newtons heres T 2 and T 1 so T 2 is 294 Newtons and at an angle of 30 T 2 X is going to be 294 cosine 30 so thats gonna be 250 4.6 now t2y thats 294 sine 30 which is 147 Newtons now t1 thats 509 point to team when X is gonna be 500 9 points you x cosine 60 and thats two hundred fifty four point six Newtons and then t1 why thats 509 point 2 sine 60 which is 440 point nine eight or you could round it to 441 just to keep things simple so well need to do to make sure our answers are correct is to make sure all the forces in the X and the y direction they add up to zero lets focus on the x-direction this one is in the positive x-direction so its positive 254 and this one is in a negative x direction so these two these two forces they cancel out they add up to zero now lets focus on the forces in the y-direction these two are going in a positive y-direction so lets put a positive sign in front of them and this one is gone in the negative y-direction so if we add these two numbers 441 plus 147 thats 588 which balances out the downward force of 588 so all the forces in the X and in the Y Direction out of to 0 so which means that these two answers are indeed correct so its just a simple way for you to check your work to make sure if your numbers make sense so this is gonna be the last problem calculate the tension in the two ropes shown below now this problem is a lot easier than the last one I think its much more simpler but lets identify all the forces we have so we got the downward weight force we got t2 which is directed towards the right we have t1 which has an X component t1 x and a
t1 line so lets start with the forces in the y-direction this force is upwards so its going to be positive t1 Y this force is downward so its negative W and those are the only two forces directly in the Y direction and because the object is at rest the net force in the Y direction is zero so the weight force if we add W to both sides the weight force is equal to team on line and T 1 Y is T 1 times sine theta and theta 60 now W the wave force is mg so now we can plug in everything we need to calculate T 1 the mass is 100 the gravitational acceleration is 9.8 and then we have sine 60 so 100 times 9.8 is 980 and sine 60 is point eight six six so nine eighty divided by point eight six six thats, 11 thirty one point six but Im going to round it to the nearest whole number so Im going to say is about 1132 Newtons so thats the tension in this rope thats the value of t1 so now lets calculate t2 so lets focus on the forces in the x-direction now t2 is a positive force its directed in a positive x direction so were just going to write that t2 team 1x is in a negative x direction so Im gonna put a negative sign in front of it and this is going to be zero because the object is in equilibrium so if we add t1 x to both sides you can see that team on X is equal to t2 so t2 is equal to t1 X and we note C 1 X is t1 cosine theta t1 is 11:32 which we have down here and beta is 60 now cosine 60 is 1/2 so this is 1132 times 0.5 so half of 1132 thats 566 and so now we have the two answers but just like we did before lets check the work lets make sure all the numbers make sense so the downward weight force is a hundred times 9.8 so thats 980 Newtons t2 is 566 Newtons t1 x thats 1132 cosine 60 so thats five hundred sixty-six going this way and t1 why thats 1132 sine 60 which is 980 Newtons so the fact that the Y components or the forces in the Y Direction matches thats good that means that one under my track and the forces in the X direction they cancel as well this is positive thats negative this is negative thats positive so all of the forces in the X and in the Y Direction add up to zero so the object remains at rest is in equilibrium which means that these two answers are correct so now you know how to solve common tangent problems in physics so keep them my intention is simply a force that acts along a rope thats it so whenever a rope transmits a force typically by a pullin action and that force acting through the rope is the tension force
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