Question 7.8: A rectangular block of mass 3 kg is placed on a slope as sho...
A rectangular block of mass 3 kg is placed on a slope as shown. The angle α is gradually increased. What happens to the block, given that the coefficient of friction between the block and slope is 0.6?
Check for possible sliding
Figure 7.42 shows the forces acting when the block is in equilibrium.
Resolve parallel to the slope: F = 3g sin α
Perpendicular to the slope: R = 3g cos α
When the block is on the point of sliding F = μR so
3g sin α = μ × 3g cos α
⇒ tan α = μ = 0.6
⇒ α = 31°
The block is on the point of sliding when α = 31°.
Check for possible toppling
When the block is on the point of toppling about the edge E the centre of mass is vertically above E, as shown in figure 7.43.
Then the angle α is given by:
tan α = \frac{0.4}{0.8}
α = 26.6
The block topples when α = 26.6°.
The angle for sliding (31°) is greater than the angle for toppling (26.6°), so the block topples without sliding when α = 26.6°.
