Equilibrium and gravity on a uniform beam

[alexandertg6]

Homework Statement

a uniform beam of length 12.0 m is supported by a horizontal cable and a hinge at angle θ = 70° with the vertical. The tension in the cable is 500 N. Find the gravitational force on the beam in unit vector notation and the force on the beam from the hinge in unit vector notation.

Homework Equations

force from gravity = mg
vertical force on beam from hinge = mg
Horizontal component of gravity = 0
horizontal component of force of hinge on beam = 500N

The Attempt at a Solution

I know that the horizontal force on the beam has to be in equilibrium with the tension of the cable.

T = .5(horizontal distance) ( gravity) (mass)/ vertical distance
500 = (.5(9.8)(12sin70)m)/(12cos70)
m= 37.13
mg = 363 but its wrong =\

 

[tiny-tim]

Welcome to PF!

Hi alexandertg6! Welcome to PF!

a uniform beam of length 12.0 m is supported by a horizontal cable and a hinge at angle θ = 70° with the vertical. The tension in the cable is 500 N. Find the gravitational force on the beam in unit vector notation and the force on the beam from the hinge in unit vector notation.
…
T = .5(horizontal distance) ( gravity) (mass)/ vertical distance
500 = (.5(9.8)(12sin70)m)/(12cos70)

hmm … you've taken moments about the hinge …

but what's 12sin70º ?

 

[nlightner]

I would like to point out that the method used to find the gravitational force on the beam is not entirely correct. While it is true that the horizontal force on the beam must be in equilibrium with the tension in the cable, the equation used to find the mass of the beam (m=37.13) is not applicable in this scenario.

To find the gravitational force on the beam, we need to consider the vertical and horizontal components separately. The vertical component of the gravitational force is given by Fg = mg, where m is the mass of the beam and g is the acceleration due to gravity (9.8 m/s^2). In this case, the vertical component of the gravitational force would be Fg = (37.13 kg)(9.8 m/s^2) = 363 N.

The horizontal component of the gravitational force is equal to zero, as gravity acts only in the vertical direction. Therefore, the total gravitational force on the beam can be expressed in unit vector notation as Fg = 363 N (0i + 1j), where i and j represent the horizontal and vertical unit vectors, respectively.

As for the force on the beam from the hinge, we can use the principle of moments to calculate it. The force from the hinge can be represented as Fh = Fsinθ, where F is the tension in the cable and θ is the angle between the cable and the vertical. In this case, the force on the beam from the hinge would be Fh = (500 N)sin70° = 475 N.

In unit vector notation, this force would be expressed as Fh = 475 N (cos70°i - sin70°j). It is important to note that this force acts in the opposite direction of the gravitational force, as it is keeping the beam in equilibrium. Therefore, the total force on the beam from the hinge can be expressed as Fh = -475 N (cos70°i - sin70°j).

In conclusion, to accurately calculate the gravitational force and the force on the beam from the hinge, we need to consider the vertical and horizontal components separately and use the appropriate equations.