Calculating torque about a point with given forces

[Nova_Chr0n0]

Homework Statement

What is the net torque around point A?

Relevant Equations

τ = rF

I've inserted a photo of the figure, hope everyone can see it.

SOLUTION:
1. I first solved for the angle of 100 N and 50 N since I need the force that is perpendicular to point A.
>> Angle of 100 N
theta = arctan(3/4)
theta = 36.870 degree
>> Angle of 50 N
theta = arctan(12/5)
theta = 67.380 degree
2. Solving for the net torque on point A where counter-clockwise is + and clockwise is -
τ = rF
τ_A = -1[100sin(36.870)] - 1.75(70) - 2[50sin(67.380)]
τ_A = -274.808 Nm

My final answer for the torque at point A is -274.808 Nm, but when I check the answer in the textbook, it is -295 Nm. I am confused about what part I did wrong. Also, I'm new to the forum, so if my format is not understandable, kindly inform me about it. Thanks!

 

[haruspex]

Finding the angles was rather unnecessary since you are given the right triangle ratios. You could just write 100N x 1m x (3/5) etc.
But I get the same answer as you do.

 

[hmmm27]

Well, I've got 274.8 as well, though my solution is somewhat less convolute than yours.

Is there anything else to the question ?

 

[Nova_Chr0n0]

There was no additional information about the problem; I just found a worksheet pdf file to practice torque problems. I guess there was just an error in the answer. If you're curious about it, here is the link:

https://www.npsd.k12.nj.us/cms/lib04/NJ01001216/Centricity/Domain/474/Torque WORKSHEETS 2014.pdf

Well, I've got 274.8 as well, though my solution is somewhat less convolute than yours.

Is there anything else to the question ?

 

[Nova_Chr0n0]

Finding the angles was rather unnecessary since you are given the right triangle ratios. You could just write 100N x 1m x (3/5) etc.
But I get the same answer as you do.

I guess there was just an error in the given answer for the file that I've encountered. Also, thanks for letting me know a solution with less work when involving triangle ratios. Really appreciate it!

 

[Lnewqban]

I guess there was just an error in the given answer for the file that I've encountered. Also, thanks for letting me know a solution with less work when involving triangle ratios. Really appreciate it!

Welcome, @Nova_Chr0n0 !

That dimension of 1 m to the right of the figure suggests that either a force that is applied at the right end, or the weight of the 3-meter bar, is missing.