# A block with a mass of 5.0 kg is held in equilibrium on an incline of an angle with a measure of 30.0 degrees by the horizontal force F. Find the magnitude of F. Find the normal force exerted by the inbline on the block. (disregard friction)

Well, I don't even know how to start. I DO know I have to use trig at some point... but help??

img145.imageshack.us/img145/8909/trigi6.png

so I just use these equations to find each side and solve for F?

F as a horizontal force, has components up the plane, and normal to the plane. The component up the plane is Fcos30. Then normal to the plane is Fsin30. The mass has part of its weight in the dirction normal to the plane (mgcos30), and up the plane mgsin30.

So, the forces in the plane direction is
Fcos30 and mgsin30. They sum to zero, so find F.
Fcos30=mgSin30
F =-mg tan30
check my thinking.
sorry for the error before.

You should try to reproduce the answer yourself. You can also decompose the normal force in the horizontal and the vertical direction. If you set the total force in these directions zero you get:

F_{n}cos(30°) - Mg = 0

F_{n}sin(30°) + F = 0

28.3 is what I got when I did the last equation of yours...

and when I did mgSin30 / cos30.

is this correct?

I'm sorry, but I don't quite understand what you mean...

and what does the _{n} mean in those equations?

F_{n} is the normal force.

Let'S start from the beginning. Accordingto Newton's second law:

F = m a

If a mass is not accelerating, then the total force acting on it must be zero. In this case there are three forces acting on the mass:

Gravity acts downward with a force of Mg

A force of F is exerted in the horizontal direction

A normal force F_{n} is exerted by the incline on the mass in the normal direction. The normal direction is at 30 degrees from the verical.

This means that the component of the normal foce in the vertical direction is

F_{n}cos(30°)

and in the horizontal direction it is

F_{n}cos(30°)

The component of the gravitational force in the vertical direction is -Mg, the minus sign means that it is directed downward.

The total force in the vertical direction is thus:

F_{n}cos(30°) - Mg

and in the horizontal component of the total force is:

F_{n}Sin(30°) + F

Count: This is the wrong approach to the problem. She needs to follow the directions I gave. Mg may be the vertical force, but on a plane she needs the normal and component parallel to the plane.
Thanks.

Yes.

## To find the magnitude of the horizontal force F, we need to set up the equation for equilibrium in the horizontal direction. The only force acting in the horizontal direction is the component of the weight of the block along the incline, which is given by:

F = mg * sin(30°)

Plugging in the values, we get:
F = (5.0 kg) * (9.8 m/s^2) * sin(30°)
F = 24.5 N

So the magnitude of the horizontal force F is 24.5 N.

To find the normal force exerted by the incline on the block, we need to set up the equation for equilibrium in the vertical direction. The forces acting in the vertical direction are the weight of the block and the vertical component of the normal force. The weight of the block is mg, where g is the acceleration due to gravity, which is approximately 9.8 m/s^2. The vertical component of the normal force is given by:
Fn = mg * cos(30°)

Plugging in the values, we get:
Fn = (5.0 kg) * (9.8 m/s^2) * cos(30°)
Fn = 42.4 N

So the normal force exerted by the incline on the block is 42.4 N.

## To find the magnitude of the horizontal force F, you can use the equation Fcos30 = mgsin30. This equation states that the horizontal component of the force F (Fcos30) is equal to the gravitational force (mg) acting on the block parallel to the incline.

To find the normal force exerted by the incline on the block, you can use the equation Fsin30 + Fn = 0. This equation states that the sum of the vertical component of the force F (Fsin30) and the normal force (Fn) is equal to zero since the block is in equilibrium.

To solve for F and Fn, you can substitute the values of mg and sin30 into the equations:

Fcos30 = mgsin30

Fsin30 + Fn = 0

Simplifying the first equation, you get:

F = (mgsin30) / cos30

Next, you can substitute the value of F into the second equation:

(mgsin30) / cos30 * sin30 + Fn = 0

Simplifying further:

(mgsin30)^2 / (cos30 * sin30) + Fn = 0

Solving for Fn:

Fn = - (mgsin30)^2 / (cos30 * sin30)

You can then substitute the given mass (5.0 kg) and solve for Fn:

Fn = - (5.0kg * 9.8m/s^2 * sin30)^2 / (cos30 * sin30)