Tension and normal force on a car held by a cable on a frictionless ramp

[saturn67]

1200kg car is held in place by a light cable on a very smooth (frictionless) ramp, as shown in the figure View Figure . The cable makes an angle of 31.0 dgree
above the surface of the ramp,
and the ramp itself rises at 25 degree above the horizontal.

how to find tension of the cable and How hard does the surface of the ramp push on the car?

please help

 

[learningphysics]

Divide the forces into components along the ramp, or perpendicular to the ramp...

If the x-axis is along the ramp... y - axis is perpendicular to the ramp... write the equations:

$$\Sigma{F_x} = 0$$
$$\Sigma{F_y} = 0$$

 

[saturn67]

what angle should i use the 25 degree rite?

Fx=T-11760sin(25)=0
Fy=T-11760cos(25)=0

 

[learningphysics]

what angle should i use the 25 degree rite?

Fx=T-11760sin(25)=0
Fy=T-11760cos(25)=0

What is the component of T along the ramp and perpendicular to the ramp... remember the cable makes an angle of 31 degrees with the ramp.

Also, don't forget the normal force in the Fy direction.

 

[saturn67]

so Fy for nomal force = 11760cos(25) + 11760cos(31)?

 

[learningphysics]

so Fy for nomal force = 11760cos(25) + 11760cos(31)?

No... just modify these equations that you already have:

Fx=T-11760sin(25)=0
Fy=T-11760cos(25)=0

you need to just change a few things... you need to include normal force as one of the forces in the y-direction... and you also need to change from T to the component of T in the x and y directions...

 

[saturn67]

Fy=T-11760cos(25)+n=0
Tx=11760*cos(31)
Ty=11760*sin(31)

i'm not so sure how to get component for T :(

 

[learningphysics]

Fy=T-11760cos(25)+n=0
Tx=11760*cos(31)
Ty=11760*sin(31)

i'm not so sure how to get component for T :(

you got it...

Tx=11760*cos(31)
Ty=11760*sin(31)

now use these in your Fy and Fx equations instead of just T. you've inserted the normal force correctly.

 

[saturn67]

ramp push on the car is normal force right? humm i still got the wrong answer
n= -11760sin(31)+11760cos(25)
n=4601.33

 

[learningphysics]

ramp push on the car is normal force right? humm i still got the wrong answer
n= -11760sin(31)+11760cos(25)
n=4601.33

no that's not right... you need to get the equations right first... you're almost there... post the Fx and Fy equation... but remember to use Tx, Ty...

post the equations...

 

[saturn67]

Fx=11760cos(31)-11760sin(25) = 0
Fy=11760sin(31)-11760cos(25)+n = 0

right?

 

[learningphysics]

Fx=11760cos(31)-11760sin(25) = 0
Fy=11760sin(31)-11760cos(25)+n = 0

right?

no.

these are the equations you need:

Fx=Tcos(31)-11760sin(25) = 0
Fy=Tsin(31)-11760cos(25)+n = 0

now, you have 2 equations and 2 unknowns (T and n). solve for T and n.

 

[saturn67]

oooo so u don;t need weight for tension in component form

 

[saturn67]

Thank you a lot learning physics :D

 

[learningphysics]

oooo so u don;t need weight for tension in component form

I don't understand what you mean...

 

[learningphysics]

Thank you a lot learning physics :D

You got the answers? Cool! Good job!

 

[saturn67]

w=mg
Fx= Tx-w*sin(25)
since you said Tx= T*cos(31)

so T don't has any w in it?

 

[learningphysics]

w=mg
Fx= Tx-w*sin(25)
since you said Tx= T*cos(31)

so T don't has any w in it?

Yeah. Tx = Tcos31 and Ty = Tsin31. Were you able to get the answers?

 

[saturn67]

yea
i got right answer
thank you