Question

f4 - 2. in each case, determine the moment about point o. problem f4 - 2

Explanation:

Step1: Resolve the force into components

The force $F = 500$ N. The vertical component $F_y=500\times\frac{3}{5}=300$ N and the horizontal component $F_x = 500\times\frac{4}{5}=400$ N.

Step2: Calculate moment for part (a)

For part (a), using the principle of moments $M_O=\sum r\times F$. The vertical - component creates a clock - wise moment and the horizontal component creates a counter - clockwise moment.
The moment due to the vertical component $M_{Oy}=300\times(3 + 1)=1200$ N·m (clock - wise).
The moment due to the horizontal component $M_{Ox}=400\times1 = 400$ N·m (counter - clockwise).
The net moment $M_O^a=1200-400 = 800$ N·m (clock - wise).

Step3: Calculate moment for part (b)

For part (b), the vertical component $F_y = 300$ N creates a counter - clockwise moment and the horizontal component $F_x=400$ N creates a counter - clockwise moment.
The moment due to the vertical component $M_{Oy}=300\times(2 + 1)=900$ N·m (counter - clockwise).
The moment due to the horizontal component $M_{Ox}=400\times(1 + 1)=800$ N·m (counter - clockwise).
The net moment $M_O^b=900 + 800=1700$ N·m (counter - clockwise).

Answer:

(a) $M_O = 800$ N·m (clock - wise); (b) $M_O=1700$ N·m (counter - clockwise)