# Tension Force Physics Problems, Two Ropes or Cables on Hanging Mass With Angles, Static Equilibrium

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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   