Question

I used a 7.5 kg wheelbarrow to carry a 35 kg rock. If the rock is placed with its centre of gravity 0.2 m in front of the wheel and the centre of gravity of the wheelbarrow is 0.1 m in front of the wheel what force must be applied to the handle 0.6m from the wheel to keep the wheelbarrow horizontal?

Answer 1

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Answer on Question #64418, Physics / Mechanics | Relativity

I used a $7.5\mathrm{kg}$ wheelbarrow to carry a $35\mathrm{kg}$ rock. If the rock is placed with its centre of gravity $0.2\mathrm{m}$ in front of the wheel and the centre of gravity of the wheelbarrow is $0.1\mathrm{m}$ in front of the wheel what force must be applied to the handle $0.6\mathrm{m}$ from the wheel to keep the wheelbarrow horizontal?

Solution:

$$M_1 = m_1 \times g \times d_1 = 7.5 \, \mathrm{kg} \times 9.8 \, \mathrm{m/s^2} \times 0.1 \, \mathrm{m} = 7.35 \, \mathrm{Nm}$$ $$M_2 = m_2 \times g \times d_2 = 35 \, \mathrm{kg} \times 9.8 \, \mathrm{m/s^2} \times 0.2 \, \mathrm{m} = 68.6 \, \mathrm{Nm}$$ $$\sum M = M_1 + M_2 = 7.35 \, \mathrm{Nm} + 68.6 \, \mathrm{Nm} = 75.95 \, \mathrm{Nm}$$

The algebraic sum of points is zero or moment of force, rotating it clockwise moment equal force that rotates counterclockwise

$$M = Fd = F = M/d = 75.95 \, \mathrm{Nm} / 0.6 \, \mathrm{m} = 126.6 \, \mathrm{N}$$

Answer: 126.6 N