Draw a force diagram for the place where the cat burglar is hanging onto the cable. There is the weight (force due to gravity) of the cat burglar downward = Fg, and that weight (in Newtons) is given in the problem. There is a tension to the right from the horizontal (x) part of the cable, call it T1. There is another tension in the cable pulling up-and-to-the-right, call it T2.
Now make a chart with columns x (horizontal), y (vertical), and "?" (idk). Fg is in the y column. T1 is in the x column and T2 is in the "?" column. Using the angle they give you and trig find the horizontal and vertical components of T2: T2cosθ goes in the x column and T2sinθ goes in the y column.
Now that you have a chart, add the forces in the y direction using Newton's 2nd law: ∑F=T2sinθ-Fg=ma=0, and it =0 because the part of the cable where the cat burglar is holding is not accelerating (actually it isn't moving at all).
You should be able to solve for T2.
#15: When you have T2 which is the answer to #14, you can use Newton's 2nd law to find the force (tension) in the horizontal cable: ∑F=T2cosθ-T1=ma=0 (and it =0 because the point where the burglar is holding is not accelerating so a=0). You should be able to solve for T1.
2 comments:
Draw a force diagram for the place where the cat burglar is hanging onto the cable. There is the weight (force due to gravity) of the cat burglar downward = Fg, and that weight (in Newtons) is given in the problem. There is a tension to the right from the horizontal (x) part of the cable, call it T1. There is another tension in the cable pulling up-and-to-the-right, call it T2.
Now make a chart with columns x (horizontal), y (vertical), and "?" (idk). Fg is in the y column. T1 is in the x column and T2 is in the "?" column. Using the angle they give you and trig find the horizontal and vertical components of T2: T2cosθ goes in the x column and T2sinθ goes in the y column.
Now that you have a chart, add the forces in the y direction using Newton's 2nd law: ∑F=T2sinθ-Fg=ma=0, and it =0 because the part of the cable where the cat burglar is holding is not accelerating (actually it isn't moving at all).
You should be able to solve for T2.
#15: When you have T2 which is the answer to #14, you can use Newton's 2nd law to find the force (tension) in the horizontal cable: ∑F=T2cosθ-T1=ma=0 (and it =0 because the point where the burglar is holding is not accelerating so a=0). You should be able to solve for T1.
Hope that helps. Good luck.
thank you!
Post a Comment