# 226 - Moment of force about different points

**Problem 226**

In Fig. P-226 assuming counterclockwise moments as positive, compute the moment of force F = 200 kg and force P = 165 kg about points A, B, C, and D.

**Solution 226**

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$F_H = 160 \, \text{ kg}$

$F_V = F(\frac{3}{5}) = 200(\frac{3}{5})$

$F_V = 120 \, \text{ kg}$

$P_H = P(\frac{2}{\sqrt{13}}) = 165(\frac{2}{\sqrt{13}})$

$P_H = 91.526 \, \text{ kg}$

$P_V = P(\frac{3}{\sqrt{13}}) = 165(\frac{3}{\sqrt{13}})$

$P_V = 137.288 \, \text{ kg}$

**Moment of force F about points A, B, C, and D:**

$M_A = 180 \, \text{ kg}\cdot \text{m}$ → *answer*

$M_B = -6(0.3)F_H = -6(0.3)(160)$

$M_B = -288 \, \text{ kg}\cdot \text{m}$ → *answer*

$M_C = -0.3F_V - 3(0.3)F_H = -0.3(120) - 3(0.3)(160)$

$M_C = -180 \, \text{ kg}\cdot \text{m}$ → *answer*

$M_D = 5(0.3)F_V - 6(0.3)F_H = 5(0.3)(120) - 6(0.3)(160)$

$M_C = -108 \, \text{ kg}\cdot \text{m}$ → *answer*

**Moment of force P about points A, B, C, and D:**

$M_A = 0$ (this means that point A is on the line of action of force P) → *answer*

$M_B = 0.3P_V = 0.3(137.288)$

$M_B = 41.186 \, \text{ kg}\cdot \text{m}$ → *answer*

$M_C = 2(0.3)P_V + 3(0.3)P_H = 2(0.3)(137.288) + 3(0.3)(91.526)$

$M_C = 164.746 \, \text{ kg}\cdot \text{m}$ → *answer*

You can also resolve P to horizontal and vertical components at point E then take the moment of these components at point C. The answer would be the same. Try it.

$M_D = -4(0.3)P_V = -4(0.3)(137.288)$

$M_D = -164.746 \, \text{ kg}\cdot \text{m}$ → *answer*