1. moving particle with friction

Hello,

I can’t see where I am going wrong in answering this question.

The question is “ a particle with mass 9kg is sliding down a smooth inclined slope with an acceleration of 4.9ms. Find the angle of the incline”.

the smooth slope means no friction.

+x is down the slope, (which i think is all i need), so i reslove this this direction.

After resolving along F I get F = 9gsin x

Therefore I use F=ma which gives me

9gsin x = (9g)(4.9)

solving for x gives me

x = arcsin (9g X 4.9) / 9g

the 9g’s cancel giving me

x = arcsin 4.9 which can’t be solved.

Any help on where I went wrong?

2. Originally Posted by pablohacker
Hello,

I can’t see where I am going wrong in answering this question.

The question is “ a particle with mass 9kg is sliding down a smooth inclined slope with an acceleration of 4.9ms. Find the angle of the incline”.

the smooth slope means no friction.

+x is down the slope, (which i think is all i need), so i reslove this this direction.

After resolving along F I get F = 9gsin x

Therefore I use F=ma which gives me

9gsin x = (9g)(4.9)

solving for x gives me

x = arcsin (9g X 4.9) / 9g

the 9g’s cancel giving me

x = arcsin 4.9 which can’t be solved.

Any help on where I went wrong?
Your (9g)sinX = (9g)(4.9) is not right.
It should be (9g)sinX = 9(4.9)

F = ma = mass*acceleration.
9g is a force, not a mass.

So,
g*sinX = 4.9
Since g = 9.8 m/sec/sec, thenn
sinX = 4.9/9.8 = 1/2
X = arcsin(1/2) = 30 degrees ------------answer.

3. Originally Posted by pablohacker
Hello,

I can’t see where I am going wrong in answering this question.

The question is “ a particle with mass 9kg is sliding down a smooth inclined slope with an acceleration of 4.9ms. Find the angle of the incline”.

the smooth slope means no friction.

+x is down the slope, (which i think is all i need), so i reslove this this direction.

After resolving along F I get F = 9gsin x

Therefore I use F=ma which gives me

9gsin x = (9g)(4.9)

solving for x gives me

x = arcsin (9g X 4.9) / 9g

the 9g’s cancel giving me

x = arcsin 4.9 which can’t be solved.

Any help on where I went wrong?

Hello,

if the particle is accelerated at $a=4.9 \frac{m}{s^2}$ then you can calculate the accelerating force:

$F=m \cdot a \Longrightarrow F=9 kg\cdot 4.9 \frac{m}{s^2}=44.1N$

The weight of the particle is directed to the centre of the eart and can be calculated by:

$W=m \cdot g \Longrightarrow W=9 kg \cdot 9.81 \frac{m}{s^2} \approx 88.29 N$

Now $\sin(\alpha)=\frac{F}{W} \Longrightarrow \sin(\alpha)=\frac{44.1N}{88.29N} \approx 0.49949... \approx 0.5$

Therefore α = 30°

EB

4. thank you allot, i see its just that i put the mass as 9g when its measured in killograms not newtond. thank you for pointing out my silly mistake. (now that helps with the rest of the homwork.)