Help With Torque Question

[AKGAKG]

Hello everyone. I'm in Physics SL and I'm taking my paper 3 on engineering physics Monday. However I'm having trouble understanding rotational equilibrium at an angle, and I was wondering if anyone could help me with this problem: question  number 10 on this pdf document: http://www.menihek.ca/Teacher%20Pages/Paula%20Kelly_files/Physics_3204_10_11/Review_Sheet_torques.pdf

 

A uniform rod of length 2 .0 m and mass 4.0 kg is hinged at the left end.  A 25.0 kg sign is suspended from the right end.  A guy wire is connected to the end of the rod and is fastened to the wall.

 

 

1.      Determine the tension in the wire.

 

2.      Determine the vertical and horizontal components of the force acting on the hinge.

 

 

 

1.      In order to determine the tension in the wire I used the hinge as my axis of rotation  causing it to have 0 torque. In Clockwise direction the Toque would be the torque caused by the beam and the torque caused by the sign, which due to being on the same plane as the hinge would be the following: 40N X 1m X sin 90 + 250N X 2 m x sin 90= 540 NM as the torque. Counterclockwise the Torque would be caused by the perpendicular component of the tension in the wire T, which would be located at the center of the wire. The angle between the hinge and the tension would be 60 degrees as due to being located at the center. Thus the CCW torque would be T X 1.15m(which is half the length of the wire, calculated using the angel of 30 degrees and the rod length of 2 m) x sin 60. In order to find the tension in the wire 540 NM would be divided by 1.15 x sin60 giving 542 N.

 

 

 

2.      In order to determine the vertical and horizontal components of the force hinge, I saw since all the forces balance in the horizontal and vertical direction this means that since the vertical force is acting upward it's force must equal all of the forces acting down which is the mass(40 N), the sign(250 N), and the tension(542 N) which sum up to be 832 N, so it must be 832 N as well. For the horizontal force the opposite force to it the vertical component of the tension which is 542 x cos 60 which is 271 N, meaning that the horizontal force must also 271 N as well to get equilibrium. Is this the correct method, as I need to ace paper 3 to save my grade considering I royally screwed up paper 2. Please I could really use the feedback.

Edited May 7, 2016 by AKGAKG

[kw0573]

Here's a diagram of how the angle of the cross product is typically taken. Note this is the general tau = r * F * sin theta.

1. You're correct that vertical component of the tension cancels out the torque. Yet, the tension force is applied at the end of the rod instead of the center. Yet, the angle is 150°. Wire length is not important.
The systematic way would be r * T * sin (angle) = 2 (length of rod) * T * sin 150° = 540.

You got lucky that 1/(cos 30°) (mysterious half-length of string) * (sin 60°) = 1 = correct way of 2 (length of rod) * sin 150° = 1
You're thinking of doing this, which is slightly more complicated and you still need to consider rod length not string length or half-length.

torque = 2 * (T cos 60°) * sin(90°) = 540 Note that cos 60° = sin 150°. Also note that this method you need to consider two angles: the angle that vertical component of T makes with T, as well as the same vertical component makes with r. 
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For calculation of F_H. 
No you can't add 40 + 250 + 542 like that because the forces are not in the same direction. You need to add the vertical and horizontal components together from a free body diagram.
F_{H_horizontal_component} = T sin 60° = T cos 30°
F_{H_vertical_component} = T cos 60° - (weight of rod) - (weight of sign). 

Correct answer should be T = 540 N (You used invalid and meaningless quantities), F_{H_horizontal} = 468 N right, F_{H_Vertical} = 20 N down. 

Calculations conducted using g = 10 m/s². Strongly advise that you use 9.8 or 9.81 m/s².