A 1400N uniform boom at 550 to the horizontal is supported

A 1400-N uniform boom at = 55.0° to the horizontal is supported by a cable at an angle = 35.0° to the horizontal as shown in the figure below. The boom is pivoted at the bottom, and an object of weight w = 1750 N hangs from its top. (a) Find the tension in the support cable. Incorrect: Your answer is incorrect. Your response differs from the correct answer by more than 10%. Double check your calculations. kN (b) Find the components of the reaction force exerted by the pivot on the boom. horizontal component kN right vertical component kN upward

Solution

Draw a picture as follows. Use protractor and ruler if available.
Draw a 10 cm vertical line to represent the wall the hinge is attached to.
At the bottom of the vertical line, draw a 7 cm line to the right that makes a 55° angle to the ground. This line represents the boom.
From the right end of the 7 cm 55° line, draw a horizontal line back (to the left) to the vertical line.
From the right end of the horizontal line (also top, right end of the boom), draw a line 35° up and to the left until it reaches the wall (10 cm vertical line).

You will see 2 triangles, label the top point of the top triangle as Point A. This is the point where the 35° line reaches the wall.
Label the point where the cable and boom intersect as point B
Label the bottom left corner of the boom where the hinge is as Point C
Labe the point where the horizontal line intersects the wall as Point D

Draw an arrow 4 cm arrow down from Point B. This represents the 1750 N weight.
Draw an arrow 2 cm arrow down from the center of line CB.This arrow represents the weight of the boom = 1400 N
Since I do not have the length of the boom, I will let it be L.
This is a torque problem
Torque = Force * distance * sin
Both the 1750 N weight and the 1400N weight vectors make a 35° with the boom.
The cable makes a 90° angle with the boom

The pivot is the hinge

Torque = Force * distance * sin
The cable is attached to the end of the boom, so distance = L
Torque caused by tension in cable = T * L

The 1750 N weight is attached to the end of the boom, so distance = L
Torque caused by tension in cable = 1750 * L * sin 35° clockwise

The 1400 N weight of the boom is at the center of the boom so distance = L/2
Torque caused by weight of boom = 1400 * ½ * L * sin 35° clockwise

Torque counter clockwise = torque clockwise

T * L = 1400 * ½ * L * sin 35° + 1750 * L * sin 35°
L cancels out

T = 700 * sin 35° + 1750 * sin 35° = 2450 * sin 35°  = 1405.3 N = the tension of the supporting cable

b) There are 2 vertical forces that support the weight of the boom (1400 N) and the 1750 N weight, the vertical component of the tension and the vertical component of the force exerted by the hinge.

The vertical component of the tension in the cable = T * sin 35° = 806.05 N
The total weight of the boom and suspended weight =1400 + 1750 = 3150 N
So, the hinge must support 3150 – 806.05 = 2343.95 N
The vertical component of the force exerted by the hinge = 2343.95 N

There are 2 horizontal forces in this problem, the horizontal component of the tension pulling away from the wall, and the horizontal component of the force exerted by the hinge pushing against the wall. These 2 forces are in opposite directions, so let the horizontal component of the tension be positive and the horizontal component of the force exerted by the hinge is negative.

Horizontal component of the tension = T * cos 35°
Horizontal component of the force exerted by the hinge = Fh
horizontal forces = T * cos35° + Fh

Since the boom, the cable, and the hinge do not move,
horizontal forces = 0
T * cos 35° + Fh = 0

Horizontal component of the force exerted by the hinge = T * cos 35°
Horizontal component of the force exerted by the hinge = 1151.15 N