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Part 1 :
A block is at rest on the incline. The coefficients of static and kinetic friction are lrs=0.6 and lrk=0.51, respectively. The acceleration of gravity is 9.8m/s squared. What is the frictional force acting on the 46kg mass? Answer in units of N.
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Part 2 : what is the largest angle which the incline can have so that the mass does not slide down the incline? Answer in until of degrees.
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Part 3 : What is the acceleration of the block down the incline or the angle of the incline is 38 degrees? Answer in units of m/s squared.

Part 1 A block is at rest on the incline The coefficients of static and kinetic friction are lrs06 and lrk051 respectively The acceleration of gravity is 98ms s class=

Respuesta :

opudodennis

Answer:

1. 218.55 N

2. [tex]30.96^{o}[/tex]

3. [tex]2.1 m/s^{2}[/tex]

Explanation:

Part 1;

Net force [tex]F=mg sin \theta[/tex] where m is mass, g is gravitational force and [tex]\theta[/tex] is the angle of inclination

[tex]F= 46*9.8*sin 29^{o}= 218.55N[/tex]

Frictional force, [tex]F_{r}[/tex] is given by

[tex]F_{r} = \mu_{s}mg cos \theta[/tex] where [tex]\mu_{s}[/tex] is the coefficient of static friction

[tex]F_{r} = 0.6*46*9.8 cos 29[/tex]

[tex]F_{r}=236.57N[/tex]

Since [tex]F_{r}>F[/tex], therefore, the block doesn’t slip and the frictional force acting is mgh=218.55N

Part 2.

Using the relationship that

Frictional force [tex]F_{s} = \mu_{s} mg cos \theta[/tex]

[tex]mg sin \theta =\mu_{s} mg cos \theta[/tex]

[tex]\mu_{s}= \frac {sin \theta}{cos \theta}[/tex]

[tex]\mu_{s}= tan \theta[/tex]

The maximum angle of inclination [tex] \theta = tan^{-1} \mu_{s}[/tex]

[tex] \theta = tan^{-1} (0.6)[/tex]

[tex] \theta= 30.96^{o}[/tex]  

        

Part 3:

Net force on the object is given by

[tex]ma = mg sin 38 - \mu_{k} mg cos 38[/tex] where [tex]\mu_{k}[/tex] is the coefficient of kinetic friction

 [tex]a = g ( sin 38 - \mu_{k} cos 38 )[/tex]

                 = 9.8 ( sin 38 - (0.51) cos 38 )

                = [tex]2.1m/s^{2}[/tex]

Reducing the angle of the incline, increases the stability of the block on the

incline.

The correct responses are;

  • Part 1: The frictional force acting on the 46 kg, block is approximately 218.55 N.
  • Part 2: The largest angle at which the mass does not slide is approximately 30.96°.
  • Part 3: If the angle of the incline is 38°, the acceleration of the block down the incline plane is 2.095 m/s².

Reasons:

Mass of the block, m = 46 kg

Coefficient of static friction, [tex]\mu_s[/tex] = 0.6

Coefficient of kinetic friction, [tex]\mu_k[/tex] = 0.51

Acceleration due to gravity, g = 9.8 m/s²

Angle of the incline, θ = 29°

Part 1

Required:

The frictional force acting on the 46 kg. mass.

Solution:

Normal reaction of the incline plane, [tex]F_N[/tex] = m·g·cos(θ)

The maximum static frictional force = Static friction coefficient × Normal reaction

∴ Frictional force, [tex]F_{fs}[/tex] = 0.6 × 46 × 9.8 × cos(29°) ≈ 236.57

Maximum static frictional force, [tex]F_{fsm}[/tex] = 236.57 N

The force of the weight of the block acting along the incline, [tex]F_{WI}[/tex], is given as follows'

[tex]F_{WI}[/tex] = m·g·sin(θ)

∴ [tex]F_{WI}[/tex] = 46 × 9.8 × sin(29°) ≈ 218.55

Force of the weight of the block acting along the incline, [tex]F_{WI}[/tex] ≈ 218.55 N

[tex]F_{WI}[/tex] < [tex]F_{fsm}[/tex], therefore, given that the block is at rest, the frictional force acting on the block, [tex]F_{fs}[/tex], is equal too the weight of the block along the incline, [tex]F_{WI}[/tex] = 218.55 N.

The frictional force acting on the 46 kg, block [tex]F_{fs}[/tex] = 218.55 N

Part 2:

Required:

The largest angle of the incline at which the mass does not slide.

Solution:

The angle, θ, at which the mass begins to slide is given by the equation:

[tex]F_{WI}[/tex] = [tex]F_{fs}[/tex]

Which gives;

m·g·sin(θ) = m·g·[tex]\mu_s[/tex]·cos(θ)

Dividing both sides of the above equation by m·g·cos(θ), gives;

[tex]\dfrac{m \cdot g \cdot sin(\theta)}{m \cdot g \cdot cos(\theta)} = \dfrac{m \cdot g \cdot \mu_s \cdot cos (\theta)}{m \cdot g \cdot cos(\theta)}[/tex]

[tex]\dfrac{sin(\theta)}{cos(\theta)} =tan(\theta) = \mu_s[/tex]

tan(θ) = [tex]\mu_s[/tex] = 0.6

θ = arctan(0.6) ≈ 30.96°

The largest angle at which the mass does not slide, θ ≈ 30.96°

Part 3: If the angle of the incline is θ = 38°, we have;

[tex]F_{WI}[/tex] = 46 × 9.8 × sin(38°) ≈ 277.54 N

[tex]F_{fs}[/tex] = 0.6 × 46 × 9.8 × cos(38°) ≈ 213.14 N

[tex]F_{WI}[/tex] > [tex]F_{fs}[/tex], therefore, the block slides down the incline and only the kinetic friction acts.

Kinetic friction force, [tex]F_{fk}[/tex] = [tex]\mu_k[/tex]·m·g·cos(θ)

Kinetic friction force, [tex]F_{fk}[/tex] = 0.51 × 46 × 9.8 × cos(38) ≈ 181.17

Net force acting along the incline plane, F = [tex]F_{WI}[/tex] - [tex]F_{fk}[/tex]

F ≈ 277.54 - 181.17 = 96.37

The net force acting along the incline plane, F ≈ 96.37

Force = Mass × Acceleration

Therefore;

[tex]Acceleration = \dfrac{Force}{Mass}[/tex]

Which gives;

[tex]\mathrm{Acceleration \ of \ the \ block,} \ a= \dfrac{96.37\, N}{46 \, kg} = 2.095 \ m/s^2[/tex]

The acceleration of the block down the incline plane, a ≈ 2.095 m/s²

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