The drawing (which I will describe) shows a top view of a door that is free to rotate about an axis of rotation that is perpendicular to the screen. Find the net torque (magnitude and direction) produced by the forces F1 and F2 about the axis. Picture displays a yellow horizontal line (the door) whose axis is measured at 0.500m from the left side of the door. The larger portion of the door, on the right side of the axis, measures 1.10m. F1 points upward onto the very left edge of the door while F2 points at a diagonal, leaning to the left, but onto the very right edge of the door. This angle that it creates here, with a y-axis dotted line, is 30 degrees. F1 is 20 N while F2 is 35 N.
To find the net torque produced by forces F1 and F2 about the axis of rotation, we can use the equation:
τ = r1 x F1 + r2 x F2
where τ is the net torque, r1 and r2 are the respective perpendicular distances from the axis of rotation to the lines of action of F1 and F2, and x denotes the cross product.
Let's calculate the torque produced by F1 first. Since F1 points upward onto the very left edge of the door, its line of action is perpendicular to the axis of rotation. Therefore, the perpendicular distance r1 is simply the distance from the axis to F1, which is 0.500m.
τ1 = r1 x F1 = (0.500m) x (20 N)
Next, let's calculate the torque produced by F2. Since F2 points at a diagonal, leaning to the left, onto the very right edge of the door, we need to find the perpendicular distance r2.
From the given information, we know the larger portion of the door measures 1.10m on the right side of the axis. So, the total length of the door is 1.10m + 0.500m = 1.60m.
Using the triangle formed by the door, r2 can be found using trigonometry. The angle between F2 and the y-axis dotted line is given as 30 degrees. Therefore, r2 = 1.60m * sin(30°).
τ2 = r2 x F2 = (1.60m * sin(30°)) x (35 N)
Now, we can find the net torque by summing up the torques produced by F1 and F2:
τ = τ1 + τ2
Finally, calculate the magnitude and direction of the net torque using the above equation, and the given values of F1, F2, r1, and r2.
To find the net torque produced by forces F1 and F2 about the axis of rotation, we need to calculate the torque due to each force separately and then add them together.
The torque due to a force applied at a distance from the axis of rotation is given by the formula:
Torque = Force * Distance * sin(angle)
1. Torque due to F1:
The force F1 is applied at the very left edge of the door, which is a distance of 0.500m from the axis. The angle between the force and the lever arm (distance) is 90 degrees since F1 points upward and the lever arm is perpendicular to it.
Torque due to F1 = F1 * Distance * sin(angle)
= 20 N * 0.500 m * sin(90°)
= 10 N⋅m
2. Torque due to F2:
The force F2 is applied at the very right edge of the door, which is the longer portion measuring 1.10m from the axis. The angle between the force and the lever arm is 30 degrees.
Torque due to F2 = F2 * Distance * sin(angle)
= 35 N * 1.10 m * sin(30°)
= 19.126 N⋅m
3. Net torque:
To find the net torque, we add the torques due to F1 and F2 together.
Net torque = Torque due to F1 + Torque due to F2
= 10 N⋅m + 19.126 N⋅m
= 29.126 N⋅m
Therefore, the net torque produced by the forces F1 and F2 about the axis of rotation is 29.126 N⋅m. The direction of the torque is not specified in the problem statement, so the magnitude is given without direction.
Nice description.
What you have to do is this:
Sum Vertical forces and set to zero.
This the addition of the weight of the door, plus the two upward components of the F1 and F2.
sum horizontal forces and set to zero.
Now, sum moments about any point (I suggest one of the hinges), and set to zero.
These three equations will solve your system.