Help with trig/mechanics question

[NewAccount]

Can anyone help with this question? The answers given are 119.3 and 33.8 but I can’t see how to get to them

 

[Deleted member 4993]

View attachment 14163
Can anyone help with this question? The answers given are 119.3 and 33.8 but I can’t see how to get to them

Did you draw a free-body-diagram?

Break-up the applied force (Y) along the slope and perpendicular to the slope. Take weight and friction forces and normal force (perpendicular to the slope - into account through equations of equilibrium and solve.

If you are still stuck - include your FBD and rest of the work and we can help.

 

[NewAccount]

Did you draw a free-body-diagram?

Break-up the applied force (Y) along the slope and perpendicular to the slope. Take weight and friction forces and normal force (perpendicular to the slope - into account through equations of equilibrium and solve.

If you are still stuck - include your FBD and rest of the work and we can help.

I did the diagram but where do I go from there, I got R = mgcos40 + ysin40
Therefore Friction = 0.3(mgcos40 + y sin 40) but that doesn’t help me with what y is

 

[Deleted member 4993]

I did the diagram but where do I go from there, I got R = mgcos40 + ysin40
Therefore Friction = 0.3(mgcos40 + y sin 40) but that doesn’t help me with what y is

Assume the forces are concurrent (acting at a point - particle

Apply the conditions for static equilibrium.

What are the forces in x-direction (assume positive x direction to be up the slope)?

Y - 0.3*R -(8*9.81)*sin(40) = 0...............................................(1)

What are the forces in y-direction (assume positive y direction to be normal to the slope)?

R - (8*9.81)*cos(40) = 0 .........................................................(2).................. edited

You have two equations and two unknowns (Y and R) - solve!

 

[Steven G]

Please rotate the picture

 

[NewAccount]

Assume the forces are concurrent (acting at a point - particle

Apply the conditions for static equilibrium.

What are the forces in x-direction (assume positive x direction to be up the slope)?

Y - 0.3*R -(8*9.81)*sin(40) = 0...............................................(1)

What are the forces in y-direction (assume positive y direction to be normal to the slope)?

R - (8*9.81)*sin(40) = 0 .........................................................(2)

You have two equations and two unknowns (Y and R) - solve!

Then R = 78.4sin40 = 50.39
Y = (0.3x50.39) + 50.39
Y = 15.12 + 50.39
Y = 65.51
And that isn’t the answer?
Or have I done something wrong

 

[Deleted member 4993]

Then R = 78.4sin40 = 50.39
Y = (0.3x50.39) + 50.39
Y = 15.12 + 50.39
Y = 65.51
And that isn’t the answer?
Or have I done something wrong

In response #2 - I wrote the 2nd equation wrong _ I have edited it. Please fix your calculations accordingly.

Sorry about that.

Hey Otis - are you still in the corner? Need some company?

 

[NewAccount]

In response #2 - I wrote the 2nd equation wrong _ I have edited it. Please fix your calculations accordingly.

Sorry about that.

Hey Otis - are you still in the corner? Need some company?

I’m sorry I’m still confused, that still isn’t giving me the given answer?

 

[Deleted member 4993]

Scratch my response #2 - I gave you solution when the force is parallel to the plane.

I am getting in hurry and making "intelligent" mistakes! I'll post correct solution to-morrow.

In the meantime I am retracting my posts.

 

[NewAccount]

Ok thank you

 

[Deleted member 4993]

I’m sorry I’m still confused, that still isn’t giving me the given answer?

No wonder - I did it wrong see response #9. I'll fix it to-morrow.

 

[Deleted member 4993]

Assume the forces are concurrent (acting at a point - particle

Apply the conditions for static equilibrium.

What are the forces in x-direction (assume positive x direction to be up the slope)?

Y*cos(40) - 0.3*R -(8*9.81)*sin(40) = 0...............................................(1).................. edited

What are the forces in y-direction (assume positive y direction to be normal to the slope)?

-Y*sin(40) + R - (8*9.81)*cos(40) = 0 .........................................................(2).................. edited

You have two equations and two unknowns (Y and R) - solve!

This should (will) give you correct solution for impending motion "up the hill".