What is the resultant torque about the hinge?

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Homework Help Overview

The discussion revolves around calculating the resultant torque about a hinge, given two forces: a counterclockwise (CCW) force of 280N and a clockwise (CW) force of 400N. Participants reference a diagram that includes additional data relevant to the problem.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss methods for calculating torque using the formula M = F * perpendicular distance D. There are attempts to determine the angles and distances involved for both forces. Questions arise regarding the angles between the forces and the lever arms, as well as the correct units for torque.

Discussion Status

Some participants are clarifying their understanding of torque calculations and the importance of using the correct units. There is a recognition of differing results among participants, with one individual confirming a torque value that aligns with another's calculation. The discussion is ongoing, with participants seeking to verify their approaches and assumptions.

Contextual Notes

There is mention of a diagram that is pending approval, which may contain critical information for the calculations. Participants also note the importance of ensuring that distances are converted to meters for accurate torque calculations.

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Homework Statement


What is the resutant torque about the hinge?

known data: CCW force 280N, CW force 400, see diagram attached for all data. All red drawings are done by me. Black is the original problem.

Homework Equations



M = F * perpendicular distance D.

The Attempt at a Solution


Ok. This was my method.
For the right side of the hinge I drew the equivalent angle of 40 deg to the 30 cm bar.
30cm cos40deg gave me the perpendicular distance of 22.96 cm

I performed the same thing on the left side of the hinge. I cut the 90deg in half and drew a line to the line of sight of the hinge. I assumed it was half of 90.
60cos45deg gave me 42.426cm

so now Mhinge = 400N*0.2296m - 280N*0.4243m
= -26.964N
So the resultant torque is a CCW force of 26.964 N.
My buddy got an answer of -90.87 N.

Thanks in advance.
 

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I can't see the drawing yet, but I can say that it doesn't seem that the net torque can be as much as 90 N-m. (notice the correct unit for torque).

Other than that we have to wait for the picture to be approved.
 
here is a link incase the attachment doesn't work. http://img402.imageshack.us/img402/1069/s4bv6.jpg
 
Last edited by a moderator:
So you want the angle between the force and the lever arm. (Defining ccw positive)

M_A = F*d*sin90 = F*d

M_B = -F*d*sinØ

what is the angle for the clockwise torque?
 
i don't follow what you have done.
can you see my diagram?
 
Yeah I can see the url one. Your 280N torque is perpendicular to your lever arm, so the cross product will simply give M_a = F_a*d_a since the sine of 90 degrees is 1.

The other one, the one that you should give a shot at is opposite to the 280N torque (what I have called M_a), but the force is not perpendicular to the lever arm, so you have to calculate the sine of whatever that angle is.

Also, be careful to get the right units, you have cm in your drawing, but you will want meters for your torque.
 
ok i am getting now -90.87 N-m. same as my buddy.

For clarification, a force must be perpendicular to the point you are analzying and the distance between the force and point is the perpendicular distance you need when doing the M calculations?
 
Yes, that is correct.
 

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