# Tension in two strings and different angles

by Todd88
Tags: angles, strings, tension
 P: 22 1. The problem statement, all variables and given/known data In a rescue, the 72.0 police officer is suspended by two cables, as shown in the figure below. 2. Relevant equations T = mgsin(48)??? T2 = mgsin(35)??? 3. The attempt at a solution I used the above equations and it is not right. Me and my friend have been working on this for a while now and it's starting to get frustrating! Any help is appreciated!
 PF Patron HW Helper P: 3,394 You need to make a force (free body) diagram for the man. He has two tension forces pulling in the direction of those ropes and there is another force on him that you undoubtedly already know. Separate each of the 3 forces into their horizontal and vertical components. Then write that the sum of the horizontal forces is zero and the sum of the vertical forces is zero (because there is no acceleration in either direction). You should be able to solve the system of 2 equations for the two unknown tensions.
 P: 22 Okay so Fg = 705.6 which means the normal force would have to be 705.6? Since we know that, the y components of both angled strings are 705.6 but the x components are different? Correct me if I'm wrong...
PF Patron
HW Helper
P: 3,394

## Tension in two strings and different angles

Yes, Fg looks good. But there is no "normal" force. A normal force presses two objects together as in a block on a ramp.

So you have Fg down and you have T1*sin(35) up and also T2*sin(48) up.
You get this by drawing the T1 force vector at its angle, then drawing a horizontal plus a vertical arrow that ends in the same place. Use trig to find their lengths. Finally, forget the T1 and think only of the horizontal and vertical forces instead.

Sum those and set the total equal to zero.

Then do the horizontal forces.
 P: 22 T1 and T2 would be different values correct? So I would have to do a system of equations first solving for either T1 or T2 and then plug those in?
 PF Patron HW Helper P: 3,394 Yes.
 P: 22 Wait you edited your answer and now it's confusing. For the y components, wouldn't it just be 705.6? The x components would be different for both angles but how would they even out to 0?
 PF Patron HW Helper P: 3,394 It is true that the two vertical tension components add up to mg (705.6). That gives you one equation relating T1 and T2 (with two sines in it, of course). You need another equation so you can solve a system of 2 equations to find the 2 unknowns. That other equation is Sum of horizontal forces = 0.

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