Calculating the Tension in Cables for Elevator Movement

[hayley123]

a hevy piece of macinery(85kg) is being moved by elevator between floors in a building. Two identical small cranes are used. a cable connects each 'crane' to a mounting hook on the top of the piece of equipment. the cable from one crane is at an angle of 22degrees below the horizontal while the cable from the other hangs at 40 degrees. find the tension in each cable if the elevator is accelerating upwards at a constant 0.2ms^-2. and i need to include a vector diagram

Homework Equations

im thinking ticos22-t2cos40-0
ti=N/sin22something

The Attempt at a Solution

i drew a picture of what i thought it might look like and I am honestly so baffled.

 

[PhanthomJay]

The equipment is being held up by 2 cables at the angles noted. It is not in contact with the floor. The horiz comp of each cable tension force acting on the equipment uses the cos function as I think you were addressing. There's no acc in x direction, so Newton 1 applies in that direction. Now in the vert direction, what are the vert comp of each tension force, and what other force acts in the vert direction? Use Newton 2 in that dir. Solve the resulting equations.

 

[hayley123]

omg I am still confused haha but ill go through what you said step by step and see how i go thanks

 

[hayley123]

ok so far i did f=ma
F=85x9.8-0.2m/s-2
F=812N
so that's the force acting on the cables yeah?

 

[hayley123]

108N i mean

 

[hayley123]

ah nah I am way off i think

 

[PhanthomJay]

You need to draw a free body diagram sketch that shows all forces acting on the equipment...its weight acting down and the 2 tension forces acting upwards at the given angles. Break up each tension force into its x and y components. Note directions.

In the x direction, apply Newton 1, and in the y direction, apply Newton 2. You are not correctly using Newton's laws. In the y direction, for example, Newton 2 says that the net force in that direction is equal to product of the mass times its acceleration in that direction. The net force in the y direction is the algebraic sum of the weight and vert comps of the tension forces. Set that net force equal to ma... i don't see any equal signs in your incorrect equation. You then have to look in the x direction as well to get another equation (using F_net = 0 in that direction), and solve the 2 simultaneous equations for the unknown tension forces.

 

[hayley123]

ah ty ty ty think I've got it now thanks heaps for your help.