Tension & Force

Question

A uniform beam 3.60 m long and weighing 2100 N carries a 3850 N weight 1.50 m from the far end, as shown in the figure below (Figure 1). It is supported horizontally by a hinge at the wall and a metal wire at the far end.

a. How strong does the wire have to be? That is, what is the minimum tension it must be able to support without breaking?

b. What is the horizontal component of the force that the hinge exerts on the beam?

c. What is the vertical component of the force that the hinge exerts on the beam?

here is the image for the figure: https://freeimage.host/i/mWg23P

Answer 1

External forces acting on the beam are the beam weight, the cable tension (T), the force (R) exerted by the wall at the hinge (with direction of R unknown), and the 3850-Newton weight 1.50 meters from the right end of the beam.

Resolve T and R into horizontal and vertical components and apply the first condition for equilibrium:

(1) ∑F_x = Rcos θ − Tcos 30° = 0

(2) ∑F_y = Rsin θ + Tsin 30° − 3850 N − 2100 N = 0

R, T, and θ are all unknown, so a solution cannot be had for these expressions alone; there must be as many simultaneous equations as there are unknown variables.

To meet the condition for rotational equilibrium, let the axis for the torque equation pass through the hinge at O where the hinge meets the wall. This is convenient since R and the horizontal component of T both have a lever arm of zero (which means zero torque) about this hinge.

Use the convention for torque sign about an axis (+ for upward torque and - for downward torque) and note lever arms as:

3850 N --- 2.10 m from O

2100 N --- 1.80 m from O

Tsin 30° --- 3.60 m from O

Then write ∑τ₀ = (Tsin 30°)(3.60 m) − (3850 N)(2.10 m) − (2100 N)(1.80 m) = 0 to gain T as 6591.666667 Newtons (the minimum tension required of the wire).

Place this value of T in (1) and (2) above to reach:

Rcos θ = 5708.550787 Newtons (the horizontal component of the force of the hinge on the beam).

Rsin θ = 2654.166667 Newtons (the vertical component of the force of the hinge on the beam).

Then θ is the arctangent of 2654.166667/5708.550787 or 24.93587389°.

Finally, R is 5708.550787/cos 24.93587389° or 2654.166667/sin 24.93587389°, both equal to

6295.407277 Newtons.

Use a Force Table to keep track of details:

FORCE COMPONENT----------LEVER ARM RELATIVE TO O-------TORQUE ABOUT O (N•M)

Tsin 30°------------------------------------------3.60 m-------------------------------------3.60Tsin 30°

Tcos 30°-----------------------------------------0.00 m---------------------------------------------------0

2100 N-------------------------------------------1.80 m------------------------------------(-1.80)(2100)

3850 N-------------------------------------------2.10 m------------------------------------(-2.10)(3850)

Rsin θ--------------------------------------------0.00 m----------------------------------------------------0

Rcos θ-------------------------------------------0.00 m----------------------------------------------------0