Respuesta :
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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