A box is placed on the right hand side of a see saw, producing a-389 Nm torque (clockwise). If a second box with weight 76N is placed on the left hand side of the see saw, how far (radius) must it be placed from the pivot point so that the see saw is balanced, Le. in equilibrium? Assume that the force of the box is applied perpendicular (theta 90 degrees) to the see saw Hint remember that for equlibrium, similar to linear forces, the net torgque of the system must be equal to zero. Net torque: for equilibrium: T+T -0

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Chapter1: Units, Trigonometry. And Vectors
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**Educational Website Transcription**

A box is placed on the right-hand side of a see-saw, producing a -38 Nm torque (clockwise).

If a second box with weight 76 N is placed on the left-hand side of the see-saw, how far (radius) must it be placed from the pivot point so that the see-saw is balanced, i.e., in equilibrium? Assume that the force of the box is applied perpendicular (theta = 90 degrees) to the see-saw.

**Hint:** Remember that for equilibrium, similar to linear forces, the net torque of the system must be equal to zero.

**Net torque:**
\[ \sum \tau = \tau_1 + \tau_2 \]

**For equilibrium:**
\[ \tau_1 + \tau_2 = 0 \]

*Enter your calculations below:*

[Input Box]
Transcribed Image Text:**Educational Website Transcription** A box is placed on the right-hand side of a see-saw, producing a -38 Nm torque (clockwise). If a second box with weight 76 N is placed on the left-hand side of the see-saw, how far (radius) must it be placed from the pivot point so that the see-saw is balanced, i.e., in equilibrium? Assume that the force of the box is applied perpendicular (theta = 90 degrees) to the see-saw. **Hint:** Remember that for equilibrium, similar to linear forces, the net torque of the system must be equal to zero. **Net torque:** \[ \sum \tau = \tau_1 + \tau_2 \] **For equilibrium:** \[ \tau_1 + \tau_2 = 0 \] *Enter your calculations below:* [Input Box]
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