Solving Physics Problem: Why Fn = mgsinΘ?

[Esoremada]

There was a bonus word problem on my physics homework that i didnt know how to solve. its two masses, one on an incline plane connected tot he other hanging by a pulley. Heres a crudely drawn FBD of it.

http://sketchtag.com/KS3pmhlzgq

in the question Θ=37 m1=5kg and m2=6kg. assume no friction and find the acceleration and tension in the string. It says to use "special (picture of a triangle)" what's that mean.


I looked up how to solve it and found that to find the normal force on an incline its Fn=mgsinΘ

Can someone explain why this is? And I still haven't solved it, but once i understand that part Ill try again before getting help.

 

[Esoremada]

I think this might be the wrong section, sorry if it isedit: yeah just read the sticky. my bad >_<

 

[JHamm]

The normal force points perpendicularly to the surface; draw out the surface, the horizontal, the angle between them, the force of gravity and the normal force and try to use some geometry to get the $\sin \theta$.
Alternatively I can never remember when to use $\sin$ or $\cos$ in these problems, I just think what would happen at $0^o$ and $90^o$ (Which angle would make the force disappear) to figure it out.

 

[Doc Al]

I looked up how to solve it and found that to find the normal force on an incline its Fn=mgsinΘ

This isn't true.

To find the normal force, consider the force components on the mass perpendicular to the surface. What must they add to?

You may find this helpful: Inclined Planes

 

[asteriatic]

Hello,

Thank you for sharing your problem and FBD diagram. It seems like you are working on a classic physics problem involving forces on an inclined plane and a pulley system. Let me explain why Fn=mgsinΘ and how you can approach solving this problem.

First, let's define the variables in this problem. m1 and m2 are the masses of the objects on the inclined plane and hanging from the pulley, respectively. Θ represents the angle of the incline, which is given as 37 degrees in this case. The bonus word problem also mentions a "special triangle," which is likely referring to the right triangle formed by the inclined plane and the horizontal ground.

Now, let's focus on the inclined plane and the object with mass m1. Since the plane is at an angle, the force of gravity acting on m1 can be broken down into two components: one parallel to the plane and one perpendicular to the plane. The component parallel to the plane is m1gcosΘ, and the component perpendicular to the plane is m1gsinΘ. This is where the term mgsinΘ comes from in the equation Fn=mgsinΘ.

The normal force, Fn, is the force exerted by the inclined plane on the object. It is always perpendicular to the plane, so it cancels out the perpendicular component of gravity, m1gsinΘ. This is necessary for the object to remain in equilibrium and not slide down the incline.

Now, let's consider the object with mass m2 hanging from the pulley. The tension in the string, T, is the force that the object exerts on the string. This tension is equal to the weight of the object, which is m2g. This is because the object is not accelerating in the vertical direction, so the forces must be balanced.

To solve for the acceleration and tension in the string, you can use Newton's second law, which states that the net force on an object is equal to its mass multiplied by its acceleration. In this case, the net force is the difference between the tension in the string and the component of gravity acting on m1 parallel to the plane. This can be represented as:

m1gcosΘ - T = m1a

Using the equation T=m2g, we can substitute for T and solve for acceleration, a:

m1gcosΘ - m2g =