Solving Block on Incline Homework Problem

[shikamarufoo]

Homework Statement

A 725 gm block slides down an inclined plane that makes an angle of 30° with the horizontal. It accelerates at 1.5 m/s2.

a) Find the magnitude and direction of the total force exerted on the block by the plane. For the direction, give the angle as measured counterclockwise from the vertical (in degrees).

b)What are the magnitude and direction of the force exerted on the plane by the block? For the direction, give the angle as measured counterclockwise from the vertical (in degrees).


I figured out the force of Friction was 2465

Homework Equations

fnet = ma
Not really sure on what other equations to use

The Attempt at a Solution

For a) I tried drawing a fbd and making a a triangle to find the hypotenuse to get the magnitude of the normal force, but I'm not getting it right. I also did sqrt(NormalForce^2 + Friction^2). For b), I'm not sure where to start.

Could you please walk me through this because I'm a bit confused now. Thanks~
Can someone please walk me through this because I'm a bit confused.

 

[Delphi51]

I figured out the force of friction was 2465

Probably on the right track, but did you remember to convert the grams to kg?
I don't think you need this force because it is part of the plane on box force you are trying to find.

This is an unusual question! Back to basics, I think.
Sum of the forces on the box = ma.
Gravity force + force of plane on box = ma

It is 2D so you need to work with two perpendicular directions. You could do it with horizontal and vertical directions or go with along the plane and into the plane.

 

[thomate1]

Sure, solving this problem involves using Newton's laws of motion and some trigonometry. First, let's draw a free body diagram for the block on the inclined plane.

The forces acting on the block are its weight (mg) acting vertically downwards, the normal force (N) acting perpendicular to the plane, and the force of friction (Ff) acting parallel to the plane in the opposite direction of motion. Since the block is accelerating, there must be a net force acting on it. Using Newton's second law (Fnet = ma), we can set up the following equation:

Fnet = ma

Where Fnet is the sum of all the forces acting on the block. In this case, it is equal to the weight (mg) minus the force of friction (Ff):

Fnet = mg - Ff

Now, let's break down the forces into their components. The weight can be broken down into its x and y components, where the x component is mg sin(30°) and the y component is mg cos(30°). The normal force is equal to the y component of the weight (mg cos(30°)) since it is acting perpendicular to the plane. The force of friction is equal to the x component of the weight (mg sin(30°)).

Now, we can substitute these values into our equation for Fnet:

Fnet = (mg cos(30°)) - (mg sin(30°))

Since we know the mass (725 gm) and the acceleration (1.5 m/s^2), we can solve for the magnitude of the net force:

Fnet = (0.725 kg)(1.5 m/s^2) = 1.088 N

To find the direction of the net force, we can use trigonometry to find the angle between the net force and the vertical. This angle will be the same as the angle of the inclined plane (30°). So, the magnitude and direction of the net force is 1.088 N at an angle of 30° counterclockwise from the vertical.

Next, let's consider the force exerted on the plane by the block. By Newton's third law, we know that for every action, there is an equal and opposite reaction. This means that the force exerted on the plane by the block is equal in magnitude and opposite in direction to the force exerted on the block by the plane. So