Question

Consider a uniform cubical box of side a on a rough floor that is to be moved by applying minimum possible force F at a point b above its centre of mass (see figure). If the coefficient of friction is \( \mu \) = 0.4, the maximum possible value of 100 × \( {b \over a} \) for box not to topple before moving is .......

Hint:

Pushing at height b above the centre of mass with the minimum force \(F=\mu mg\); the box is just about to topple when the normal reaction acts at the front bottom edge. Take torques about the centre of mass (or about the front edge).

Answer 1

Correct Answer: 75

Step 1: Minimum force to start sliding: \(F=f=\mu N=\mu mg\) (N = mg).
Step 2: Torque about the centre of mass: F (at height b above it) and friction f (at a/2 below it) both tend to rotate the box forward (same sense). The normal reaction N must supply the opposite torque, so it acts at a distance x in front of the centre: \(Nx=Fb+f\dfrac a2\).
Step 3: \(mg\,x=\mu mg\left(b+\dfrac a2\right)\Rightarrow x=\mu\left(b+\dfrac a2\right)\). For no toppling \(x\le\dfrac a2\).
Step 4: \(0.4\left(b+\dfrac a2\right)\le\dfrac a2\Rightarrow b\le\dfrac a2\left(\dfrac1{0.4}-1\right)=0.75\,a\). So \(100\,\dfrac ba=75\).
Answer: 75