# Physics Dynamics Question homework help

• huzjm
In summary, the acceleration in the system of a fletcher’s trolley, with m1 = 1.0 kg and m2 = 9.0 kg and a 50 N force of friction, is approximately 3.3 m/s².
huzjm

## Homework Statement

Find the acceleration in the system of a fletcher’s trolley given m1 = 1.0 kg and m2 = 9.0 kg and a 50 N force of friction exists.
θ = 33°

See the image below, to explain the diagram. This one has an angle of 33 degrees that is why it is hard.

http://i39.tinypic.com/3497zo9.png

I am not sure.

## The Attempt at a Solution

Okay so first I found out the horizontal component of m1 by the formula cos(33)*mass(which is 1.0kg) and got the value 0.838 , then I put the formula Fnet=[(M2*A)-Ff]/(m1+m2)
which means 9.0kg*9.81m/s2 - 50 N divided by (0.838+9) .. the answer was 3.9 M/s2 but this is the wrong answer, i asked my teacher, he said you are somewhere close but this is not right. I cannot figure this out, I have been trying different things since hours.

welcome to pf!

hi huzjm! welcome to pf!

(try using the X2 button just above the Reply box )
huzjm said:
Fnet=[(M2*A)-Ff]/(m1+m2)

sorry, but even if you mean a = Fnet/(m1+m2) = [(M2*A)-Ff]/(m1+m2), that's still completely wrong

try calling the tension "T", and doing two F = ma equations (one for each block)

alternatively, if you're treating the two blocks as a single system, your m in ma has to be the total (unadjusted) mass, and you have to use the component of m1g parallel to the string

try it both ways … what do you get?

Thank you very much :) In fact, thank you really very much :)
This is what I did
Fnet = ma = 9kg * a = mg - T = 9kg * 9.8m/s² - T = 88.2N - T
→ T = 88.2 - 9a

For the lighter mass,
Fnet = ma = 1kg * a = T - mgsinΘ - Ff = T - 1kg * 9.8m/s² * sin33º - 50N
1kg * a = T - 5.34N - 50N = T - 55.34N
→ T = a + 55.34

Since T = T,
88.2 - 9a = a + 55.34
32.86 = 10a
a ≈ 3.3 m/s²

excellent!

(btw, you'll notice you could get the same result by treating it as a one-dimensional motion, with a single body with a mass of 10 kg, friction of 50 N and gravitational forces of 9g N and -gsin33° N )

Oh so the formula should have been acceleration= [(9.0kg*9.81m/s) - 50N - (9.81m/s * sin 33°)] / 10 KG

Thank you really very much :)

I mean m/s(square)

## What is Newton's first law of motion?

Newton's first law of motion, also known as the law of inertia, states that an object will remain at rest or in uniform motion in a straight line unless acted upon by an external force.

## What is the difference between velocity and acceleration?

Velocity is a vector quantity that describes the speed and direction of an object's motion. Acceleration is the rate of change of an object's velocity over time.

## How does air resistance affect falling objects?

Air resistance, also known as drag, is a force that acts in the opposite direction of an object's motion through the air. This force increases as the object's speed increases, and it can cause objects to reach a terminal velocity, where the downward force of gravity is equal to the upward force of air resistance.

## What is the difference between static and kinetic friction?

Static friction is the force that prevents two objects from sliding past each other when they are at rest. Kinetic friction, on the other hand, is the force that opposes the relative motion between two objects that are already in motion.

## How is momentum conserved in a closed system?

In a closed system, the total momentum of all objects remains constant, meaning it is conserved. This means that if one object gains momentum, another object in the system must lose an equal amount of momentum in the opposite direction.

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