A block of 150 kg mass is placed on a rough horizontal plane. The coefficient of static friction between the block and the plane is 0.33. Determine the force required to prevent the block from moving in each of the following cases:
(i) If the block is pushed by a horizontal force.
(ii) If the block is pulled by an inclined force, inclined at 20° to the horizontal.
(iii) If the block is pushed by an inclined force, inclined at 20° to the horizontal.
Answer (2)
To keep a 150 kg block stationary on a rough surface with a static friction of 0.33, a force of approximately 486.4 N is needed when pushed horizontally. For pulling or pushing at an angle of 20°, the calculations would need to adjust for vertical force components affecting normal force and friction. The general approach involves balancing forces to determine specific requirements based on the direction of applied force.
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To solve this problem, we need to calculate the minimum force required to prevent a block from moving due to static friction. The force of static friction can be calculated using the formula:
F friction = μ s ⋅ N
where:
F friction is the force of static friction,
μ s is the coefficient of static friction (given as 0.33),
N is the normal force.
The normal force ( N ) is equal to the weight of the block when the force applied is horizontal or acts vertically downwards (cases i and iii), which is calculated as:
N = m ⋅ g
where:
m is the mass of the block (150 kg), and
g is the acceleration due to gravity (approximately 9.8 m / s 2 ).
(i) Horizontal Force
For horizontal force:
Calculate the normal force: N = 150 × 9.8 = 1470 N
Calculate the force of static friction: F friction = 0.33 × 1470 = 485.1 N
Thus, a force of 485.1 N is required to prevent the block from moving.
(ii) Inclined Force (20° to the horizontal)
For a force inclined at 20° to the horizontal, the normal force changes because part of the force acts vertically:
Decompose the inclined force into its vertical component: F vertical = F ⋅ sin ( 20° )
Adjust the normal force: N = 1470 − F ⋅ sin ( 20° )
Since the block is not moving, the horizontal component of the inclined force must equal the static friction:
Solve for the horizontal force component: F ⋅ cos ( 20° ) = 485.1
You need to solve these equations simultaneously to find F .
(iii) Inclined Force (20° to the horizontal pushing down)
For a force pushing down at 20° to the horizontal, the normal force increases with the vertical component:
Decompose the inclined force into its vertical component: F vertical = F ⋅ sin ( 20° )
Adjust the normal force: N = 1470 + F ⋅ sin ( 20° )
Solve for the horizontal force component: F ⋅ cos ( 20° ) = 485.1
Similar to part (ii), you need to solve these equations simultaneously to find F .
Each case requires an understanding of the effect of the angle of the force on the normal force and static friction.
