Answer:
The answer is
The force F₂, acts on the meter stick, with a magnitude of 71.21 N
Explanation:
To solve the question, we note that the sum of moment about a point = 0 or ΣM =0
That is sum of clockwise moments = sum of anticlockwise moments
Also we are required to maintain the sign convention when calculating for the moment of the force thus
Clockwise moment = -ve and clockwise = +ve
Therefore taking moment about the axis at the 20.0 cm mark we have
-175 N × 20, 0 cm -F₁y × (80 - 20) cm = 0 where F₁y is the y component of the the force at the 80.0 cm mark
or -175 N × 20, 0 cm = F₁y × (80 - 20) cm
From the above equation, it is seen that F₁y = - (175 N × 20 cm)/(60 cm) = 58.33 N
F₁y = -58.33 N, Hence the correct force at the 80.0 cm mark is
58.33 N in the opposite direction as F₁y or F₂y
Hence the magnitude of F₂ can be found by resolving the force into components thus
y component of the force F₂ = 58.33 N angle to the vertical is given by
90 - 55 or 35 ° therefore the y which is the vertical component = 58.33 N at an angle 35 ° to the vertical, we have
F₂ × cos (35 °) = 58.33 N
∴ F₂ = (58.33 N)/(cos (35 °))
= 71.21 N
Therefore the force F₂ acts on the meter stick with a magnitude of 71.21 N
