Questions: One end of a light inextensible string is attached to a tool box of mass 2.5 kg which is lying on a horizontal table. The string passes over a smooth pulley and is tied at the other end to a bag of mass 1.4 kg. a Draw a diagram showing the forces acting on the tool box. b If the tool box is just on the point of sliding, find a value for μ, the coefficient of friction.

Solution Steps

The forces acting on the toolbox are:

• Weight (W): Acting downwards, equal to _mg_ where _m_ is the mass of the toolbox (2.5 kg) and _g_ is the acceleration due to gravity (9.8 m/s²).
• Normal Reaction (R): Acting upwards, perpendicular to the table, equal in magnitude to the weight.
• Tension (T): Acting horizontally to the right, due to the string attached to the bag.
• Friction (F): Acting horizontally to the left, opposing the impending motion. Since the toolbox is on the point of sliding, the friction is at its limiting value, _F_ = _μR_, where _μ_ is the coefficient of friction.

A diagram would show the toolbox with these four force vectors acting on it.

The forces acting on the bag are:

• Weight (W'): Acting downwards, equal to _m'g_, where _m'_ is the mass of the bag (1.4 kg).
• Tension (T): Acting upwards, due to the string.

Since the system is at equilibrium (just about to move), the tension in the string is equal to the weight of the bag: _T_ = _m'g_.

For the toolbox to be on the point of sliding, the tension in the string must equal the maximum frictional force:

_T = F_

_m'g = μR_

Since _R = mg_, we have:

_m'g = μmg_

Solving for μ:

_μ = m'/m_

_μ = 1.4 kg / 2.5 kg_

_μ = 0.56_

Final Answer: