contestada

Jason works for a moving company. A 75 kg wooden crate is sitting on the wooden ramp of his truck; the ramp is angled at 11°.
A. What is the magnitude of the force, directed parallel to the ramp, that he needs to exert on the crafe to fet it to start moving UP the ramp?
B. What is the magnitude of the force directed parallel to the ramp, that he needs to exert on the crate to get it to start moving DOWN the ramp?

Respuesta :

Answer:

Part a)

F = 501.50 N

Part b)

F = 220.7 N

Explanation:

Part a)

When wooden craft is sliding upwards then the force of friction is along the inclined plane acting opposite to the slide

so it is downward along the inclined plane

now we have

[tex]F = mg sin\theta + \mu mg cos\theta[/tex]

here we know that

coefficient of friction between two wooden surface = 0.5

now we have

[tex]F = 75(9.81) sin11 + 0.5(75)(9.81) cos11[/tex]

[tex]F = 501.50 N[/tex]

Part b)

When wooden craft is sliding downwards then the force of friction is along the inclined plane acting opposite to the slide

so it is upward along the inclined plane

now we have

[tex]F = \mu mg cos\theta - mg sin\theta[/tex]

here we know that

coefficient of friction between two wooden surface = 0.5

now we have

[tex]F = 0.5(75)(9.81) cos11 - 75(9.81) sin11 [/tex]

[tex]F = 220.7 N[/tex]

onyebuchinnaji

The parallel force that will be exerted to enable the crate move UP is 501.27 N.

The parallel force that will be exerted to enable the crate move DOWN is 220.5 N.

"Your question is not complete, it seems to be missing the following information";

The coefficient of kinetic friction, μk, is 0.5

The given parameters;

  • mass of the wooden crate, m = 75 kg
  • angle of inclination, θ = 11⁰

The parallel force that will be exerted to enable the crate move UP is calculated as follows;

[tex]F_x = F_k + Wsin(\theta)\\\\F_x = \mu_k mgcos(\theta) + W sin(\theta)\\\\F_x = 0.5 \times 75 \times 9.8 \times cos(11) \ + \ 75\times 9.8\times sin(11)\\\\F_x = 501.27 \ N[/tex]

The parallel force that will be exerted to enable the crate move DOWN is calculated as follows;

Here the weight of the box must overcome the frictional force in order to slide down.

[tex]F_x = F_k - Wsin(\theta) \\\\F_x = \mu_k mgcos(\theta) - mgsin(\theta) \\\\F_x = (0.5\times 75\times 9.8 \times cos(11)) - (75 \times 9.8\times sin(11)) \\\\F_x = 220.5 \ N[/tex]

Learn more here:https://brainly.com/question/20597467

Otras preguntas