Math Problem Statement

A uniform 11.3 N picture frame is supported as shown. Find the force FT,1 in the cord that is required to hold the frame in this position. Answer in units of N.

A uniform 11.3 N picture frame is supported as shown. Find the force FT,1 in the cord that is required to hold the frame in this position. Answer in units of N.

Solution

To determine the force FT,1F_{T,1} in the cord required to hold the frame in position, we'll analyze the system using equilibrium conditions for both forces and torques.

Step 1: Identify the forces acting on the frame

The forces involved are:

  1. The weight of the frame, W=11.3NW = 11.3 \, \text{N}, acting vertically downward at the center of the frame.
  2. The force FT,1F_{T,1} in the cord making an angle of 5050^\circ with the horizontal.
  3. The vertical force FT,2F_{T,2} acting upward at the left corner of the frame.
  4. The horizontal force FF acting at point P on the right side of the frame.

Step 2: Write the equations of equilibrium

Force Equilibrium

For the frame to be in equilibrium, the sum of all horizontal and vertical forces must be zero:

  • Horizontal forces: FT,1cos(50)=FF_{T,1} \cos(50^\circ) = F

  • Vertical forces: FT,1sin(50)+FT,2=11.3NF_{T,1} \sin(50^\circ) + F_{T,2} = 11.3 \, \text{N}

Torque Equilibrium

Let's take the torque about the point where FT,2F_{T,2} is applied (top-left corner of the frame). Assuming counterclockwise torque as positive:

  • The torque due to W=11.3NW = 11.3 \, \text{N} (acting downward at the center of the frame, 14 cm below the top and 14 cm horizontally from the left): τW=11.3N×14cm=158.2Ncm(clockwise, negative)\tau_W = 11.3 \, \text{N} \times 14 \, \text{cm} = 158.2 \, \text{N} \cdot \text{cm} \quad (\text{clockwise, negative})

  • The torque due to FT,1F_{T,1} (at 28 cm from the left corner): τFT,1=FT,1sin(50)×28cm(counterclockwise, positive)\tau_{F_{T,1}} = F_{T,1} \sin(50^\circ) \times 28 \, \text{cm} \quad (\text{counterclockwise, positive})

  • The torque due to FF at point P (which does not produce torque about the top-left corner).

Setting the sum of the torques equal to zero: FT,1sin(50)×28cm=158.2NcmF_{T,1} \sin(50^\circ) \times 28 \, \text{cm} = 158.2 \, \text{N} \cdot \text{cm}

Step 3: Solve for FT,1F_{T,1}

Solving the torque equation for FT,1F_{T,1}: FT,1=158.2Ncm28cm×sin(50)F_{T,1} = \frac{158.2 \, \text{N} \cdot \text{cm}}{28 \, \text{cm} \times \sin(50^\circ)}

Calculating the value: FT,1=158.228×0.7667.35NF_{T,1} = \frac{158.2}{28 \times 0.766} \approx 7.35 \, \text{N}

Final Answer

The force FT,1F_{T,1} in the cord required to hold the frame in position is approximately: FT,17.35NF_{T,1} \approx 7.35 \, \text{N}

Would you like a more detailed breakdown or have any questions?

Related Questions:

  1. How do you determine the value of FT,2F_{T,2} in the system?
  2. What would change in the calculations if the angle of FT,1F_{T,1} were different?
  3. How would you find the horizontal force FF acting at point P?
  4. What are the conditions for equilibrium in terms of forces and torques?
  5. How does changing the weight of the frame affect the forces in the cords?

Tip:

Always choose a point with multiple unknown forces as the pivot for calculating torque to simplify your equations!

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Math Problem Analysis

Mathematical Concepts

Statics
Torque and equilibrium
Trigonometry

Formulas

Horizontal force equilibrium: F_T,1 * cos(50°) = F
Vertical force equilibrium: F_T,1 * sin(50°) + F_T,2 = 11.3 N
Torque: Στ = 0 -> F_T,1 * sin(50°) * 28 cm = 11.3 N * 14 cm

Theorems

Equilibrium conditions for forces (ΣF_x = 0, ΣF_y = 0)
Equilibrium condition for torques (Στ = 0)

Suitable Grade Level

Grades 11-12 / College Introductory Physics