A purple beam is hinged to a wall to hold up a blue sign. The beam has a mass of mb = 6.7 kg and the sign has a mass of ms = 17.2 kg. The length of the beam is L = 2.78 m. The sign is attached at the very end of the beam, but the horizontal wire holding up the beam is attached 2/3 of the way to the end of the beam. The angle the wire makes with the beam is θ = 33.9°.
What is the tension in the wire? What is the net force the hinge exerts on the beam? The maximum tension the wire can have without breaking is

T=936 N. What is the maximum mass sign that can be hung from the beam?

2. Raise Newton's equilibrium equations that establish that for a body to be in equilibrium, the sums of its vertical, horizontal forces and moments with respect to point A must be equal to zero.

3. Use the moment equilibrium equation with respect to point A (see attached image) to find the cable tension.

4. Use the equilibrium equations of vertical and horizontal forces to find the total reaction force on the hinge.

5. Use the equilibrium equation of moments with respect to point A (see attached image) to find the mass of the sign.

Remember to use algebra correctly to find the variables.

Greetings!

Lanuel Lanuel · Answer 2 · 2022-04-07T21:41:55+02:00

Based on the calculations, the tension in the wire is equal to 449.55 Newton.

Given the following data:

Mass of beam = 6.7 kg.

Mass of sign = 17.2 kg.

Length of beam = 2.78 m.

Angle = 33.9°.

Maximum tension = 936 N.

How to calculate the tension.

Since the beam is not in motion, the forces acting on it is given by:

[tex]T=F_{hx}=0=T-F_{hx}\\\\0=\frac{L}{2} m_bg cos\theta -\frac{2L}{3} Tsin \theta-Lm_s cos\theta\\\\[/tex]

Making T the subject of formula, we have:

[tex]T=\frac{3(m_b+2m_s)g}{4tan \theta} \\\\T=\frac{3(6.7+2(17.2))9.8}{4tan 33.9} \\\\T=\frac{1208.34}{2.6879}[/tex]

T = 449.55 Newton.

How to calculate the net force.

The magnitude of the net force that the hi-nge exerts on the beam is given by:

[tex]F_h = \sqrt{(F_{hx})^2+ (F_{hy})^2} \\\\F_h = \sqrt{T^2+ (m_b+m_s)^2g^2}\\\\F_h = \sqrt{449.55^2+ (6.7+17.2)^2 \times 9.8^2}\\\\F_h=\sqrt{202095.2025+54859.0084} \\\\F_h=\sqrt{256954.2109}\\\\F_h=506.91\;Newton[/tex]

How to calculate the maximum mass sign.

Mathematically, the maximum mass sign that can be hung from the beam is given by this formula:

[tex]m_{s, max}=\frac{2T_{max}sin\theta}{3g} -\frac{m_b}{2} \\\\m_{s, max}=\frac{2 \times 936 \times sin33.9}{3 \times 9.8} -\frac{6.7}{2} \\\\m_{s, max}=\frac{1044.10}{29.4} -3.35\\\\m_{s, max}=35.51-3.35\\\\m_{s, max}=32.16\;kg[/tex]

Read more on tension here: brainly.com/question/4080400