Force needed to push an incline with a mass on it

[nagaromo]

Homework Statement

Incline with the mass of M and angle of theta. mass on the incline with the mass of m. Find the force needed to push the entire system such that the mass remains at rest relative to the incline. Everything is frictionless

Homework Equations

F=ma

The Attempt at a Solution

well my teacher has gone over the problem where a=F/(M+m) but after that, I was completely lost and we arrived with the answer of F=(m+M)gtan$$\theta$$

I'll love you if you could please explain to me how to arrive to that answer! <33 THANK YOUU

 

[xsweetox]

1. draw an FBD for m on a tilted axis x prime and y prime.
2. convert that FBD to a normal axis by splitting the force arrows into components.
3. draw an FBD for M.

1. first, you need to draw all the force arrows. You got part of the forces right in your diagram. There is Fgsin(theta) and Fg(costheta). Don't forget, there is also a normal force acting on the block perpendicular to the incline. Also, The reason why the block does not slide down the ramp is because M exerts a force F(Mm) on it. Therefore, it points in the opposite direction of Fgsin(theta).

2. Next, you need to break up all these slanted force arrows into their x and y normal axis directions. after this, you will notice that the net force in the positive y direction is F(Mm)sin(theta) - Fncos(theta). This is equal to the net force in the negative y direction, Fg, as the block does not accelerate in the y direction. The net force in the positive x direction is Fgsin(theta)cos(theta). in the opposite direction you have F(Mm)cos(theta). Fgsin(theta)cos(theta) is bigger because the block does accelerate in the x direction.

3. now you draw an FBD for M. The net force in the positive x direction is Fa, the applied force. there is a force that m exerts on M, which is slanted toward the bottom left corner. When you break up this force into the x and y perpendicular forces, you get the following: a force pointing in the negative x direction F(mM)cos(theta); a force pointing in the negative y direction F(mM)sin(theta). On top of this, you have a normal force and a force of gravity acting in opposite directions. They are not equal! Fn = Fg + F(mM)sin(theta).

the acceleration of M is equal to the acceleration of m.

am = Fnet m
= Fgsin(theta)cos(theta) - F(Mm)cos(theta)

aM = Fnet m
= Fa - F(mM)cos(theta)

F(mM) = F(Mm) (Newton's third law)

then you use the equations to solve for a.

btw, i used components and not vectors. hope this helps.

 

[xsweetox]

oops, aM= Fnet M

 

[nagaromo]

thank you! :D

 

[rwisz]

Sure, no problem! Let's break down the problem step by step.

First, we need to understand the forces acting on the system. In this case, we have the force of gravity pulling the mass m down the incline, and the force of the incline pushing back on the mass m. Since the mass m is at rest relative to the incline, these two forces must be equal and opposite in magnitude.

Next, we can use Newton's second law, F=ma, to relate these forces to the acceleration of the system. Since the mass m is not accelerating, the net force on it must be zero. So we can set up the following equation: F(incline) - F(gravity) = 0.

Now, we can plug in the relevant equations for these forces. The force of gravity is given by F(gravity) = mg, where g is the acceleration due to gravity. The force of the incline is a bit trickier to calculate, but we can use trigonometry to help us. Since the incline forms a right triangle with the horizontal ground, we can use the angle theta to find the component of the incline's force that is acting in the direction of the incline. This component is given by F(incline) = mg*sin(theta).

Plugging these values into our equation, we get: mg*sin(theta) - mg = 0. We can then solve for the force needed to push the system, F, by factoring out the common term of mg: F = mg*(sin(theta) - 1).

Finally, we can simplify this expression by using the trigonometric identity sin(theta) - 1 = -cos(theta). This gives us our final answer: F = -mg*cos(theta) = (m+M)g*tan(theta).

I hope this explanation helps! Remember, when solving problems in physics, it's important to carefully consider all the forces acting on the system and use the appropriate equations to relate them to each other. Good luck with your studies!