Inclined plane, finding normal force

[GRice40]

Homework Statement

A 1160 kg car is held in place by a light cable on a very smooth (frictionless) ramp, as shown in the figure . The cable makes an angle of 31.0 degrees above the surface of the ramp, and the ramp itself rises at 25.0 degrees above the horizontal.

Find the normal force along with the tension of the cable.

Homework Equations

Fnormal = M*A

The Attempt at a Solution

I've found the tension by taking the mass X 9.8sin(25), then taking that divided by cos(31) which gave me a tension of ~5610

However, I don't know how to go about finding the normal force

 

[kreil]

I think you will find this very informative,

http://en.wikipedia.org/wiki/Inclin...rces_acting_on_an_object_on_an_inclined_plane

 

[GRice40]

Ok, I understand that in essence, the Normal Force is going to be

Fnorm= m * a(cos theta), in this case, the theta is 25 degrees, and the mass is 1160 kg.

So:

Fnorm = 1160 * 9.8(cos(25)) = 10302.91 N

That's all fine and well, except that the answer is suppose to be 7410 N. I believe it has something to do with the relationship of the weight of the car compared to the tension, and if it's not that, then I'm completely baffled =(

 

[GRice40]

Alright, just had an epiphany. The tension is, indeed, a force in the Y direction.

So here's what I came out with:

F(up) + F(tension in the y direction) - F(down) = 0

So, F(up/normal force) = F(down) - F(tension in the y direction)

Fnorm = 1160(cos(25)) - (5610(sin(31)) = 7413.55

That number is right around the correct answer that it gave me of 7410 N.

Appreciate all the help guys!

 

[rwisz]

.Based on the information provided, the normal force can be calculated using the following equation:

Fnormal = Mgcos(θ)

Where M is the mass of the car (1160 kg), g is the acceleration due to gravity (9.8 m/s^2), and θ is the angle of the ramp (25 degrees).

Plugging in the values, we get:

Fnormal = (1160 kg)(9.8 m/s^2)cos(25 degrees)

Fnormal = 11438 N

Therefore, the normal force acting on the car is 11438 N. This force is perpendicular to the surface of the ramp and is responsible for balancing out the weight of the car, which is pulling downwards due to gravity.

It is important to note that the normal force is always equal in magnitude but opposite in direction to the force exerted by an object on a surface it is in contact with. In this case, the normal force is equal in magnitude but opposite in direction to the weight of the car. This is why the car remains in equilibrium on the ramp and does not slide down due to its weight.

In summary, the normal force acting on the car is 11438 N and the tension in the cable is ~5610 N.