Solve Torque Q1: Find Tension & F in Picture Frame

[poloboy]

1. A uniform picture frame weighing 10N is supported. The cable on the right side makes an angle of 50 with the horizontal direction. Find the tension in the cords and the magnitude of the horizontal force F required to hold the frame in the position

Homework Equations

imagine a rectangular box, the length is 0.15m and width 0.3m there is a cord hanging on straight up from the top left of the box and a cord hanging on the top right of the box at a 50 degree angle. Also there is a force F pointint to the left from the bottom right corner

The Attempt at a Solution

i believe this question is a torque question, but I'm not 100% sure how to start solving. i broke down the equation into X-component and Y-component
X-comp: fcos50 - F =0
Y-comp: fsin50 + f(left) = 10

i'm kinda stuck after this point
i know that torque = Fd but i don't know where my pivot point should be and so on can anyone help. thanks!

 

[learningphysics]

You can choose any point for the torque pivot... It's best to choose a point that makes your equations simple...

I recommend the upper right point, because then you avoid having the trig stuff in your equation.

 

[ogb p]

I would approach this problem by first identifying the known values and variables. The known values are the weight of the frame (10N) and the angle of the cable (50 degrees). The variables are the tension in the cords (f) and the horizontal force (F).

To solve for the tension in the cords, we can use the Y-component equation you have already written: fsin50 + f(left) = 10. We can simplify this by factoring out f: f(sin50 + 1) = 10. Dividing both sides by (sin50 + 1), we get f = 10/(sin50 + 1). This gives us the tension in the cords, which we can calculate to be approximately 6.33N.

To solve for the magnitude of the horizontal force, we can use the X-component equation: fcos50 - F = 0. Again, we can simplify this by factoring out f: f(cos50) - F = 0. Plugging in the value we calculated for f, we get (6.33N)(cos50) - F = 0. Solving for F, we get F = (6.33N)(cos50) = approximately 4.85N.

As for the pivot point, we can choose any point along the frame where the forces are acting. In this case, since both the tension in the cords and the horizontal force are acting at the top right corner of the frame, we can choose that as our pivot point. This means that the torque equation would be: T = Fd, where T is the torque, F is the force (4.85N), and d is the distance from the pivot point to the point where the force is acting (in this case, the distance is the width of the frame, 0.3m). We can use this equation to calculate the torque acting on the frame, which is approximately 1.45Nm.

In summary, the tension in the cords is 6.33N, the horizontal force required to hold the frame in place is 4.85N, and the torque acting on the frame is 1.45Nm.