Can anyone help me with this problem? I've uploaded the original problem and the way I solved it. The way they do it, as you can see, is a bit more involved. I did it a bit differently. I also came up with a bit of a different answer. I came up with 3200 N for T1 and for T2, I came up with 563 N. I believe I'm doing it correctly and scanned/uploaded my work so you can see it and was wondering what it is I'm doing incorrectly? Thanks!

#### Attachments

• 946.7 KB Views: 1,392
• 212.4 KB Views: 126

Related Classical Physics News on Phys.org
$T_{1y}$ isn't equal to 3150 N because there's also the vertical component of the tension of the rope that the worker holds to account for

Femme_physics
Gold Member
They solved it analytically (with sum of all forces equations)-- you tried to solve it geometrically. Both are valid methods.

williem2 is right. You determined from the get go that T1y = 3150 [N]. Why, exactly? How do you know that? It's not given to you. Not T1 nor T2. The only force you know is the force of the engine, THAT being 3150 [N]. It also doesn't make any sense for T1y to be 3150 [N], because if the engine is 3150 [N] and the force of T1y is 3150 [N], that means that the vector T2y (where the man is pulling) will cause movement in the Y axis since it has a Y component vector and nothing to resist it, as mg and T1y will cancel themselves.

To solve it geometrically-- you only need 1 triangle (since you have 3 vectors emerging from one spot). One vector is vertical (mg), the other vector, T1, is added to that vector in a 10 degree angle (that's given to you straightforwardly)-- now just figure out what's the angle that T2 is added to the mg vector? (hint: it's not 80. Draw it and you will see the answer for yourself). Now just use Law of Sines to get the correct solutions.

Last edited: