Question

a block of mass 20kg is resting on a rough horizontal plane. a horizontal force of 98N is required just to slip the block. find the value of static coefficient and angle of friction​ answer should be ans. 0.50 and tan^-1 (0.5)

Answer 1

Explanation:-

As per provided information we have :-

Mass of the body is 20 kg

Horizontal Force just to slip is 98N (fl = 98N)

To find :-

• Coefficient of static friction (μs)
• Angle of friction

As we know that,

fl = μsN

where,

• fl = limiting friction
• μs = coefficient of static friction
• N = normal reaction

N = mg

where,

• N = normal reaction
• m = mass
• g = acceleration due to gravity (9.8 m/s²)

Substituting the values,

⟹fl = μsN

⟹98 = μs (20)(9.8)

⟹98 = μs 20 (98)/10

⟹1 =( μs 20 )/10

⟹ 2μs = 1

⟹μs = 1/2

μs = 0.5

So, the coefficient of friction (μs) = 0.5

════════════════════

As we know that ,

Angle of friction is (Φ) is

⟹tanΦ = μ

⟹0.5 = tanΦ

⟹Φ = tan ^-1 (0.5)

So, the angle if friction is tan ^-1 (0.5)

═══════════════════════