M1 Statics, please help

Original post by najinaji

I'd like it if you could please check my answers & also provide help. :smile:

1)

a) Find the value of θ

12sinθ=8sin90
12sinθ=8
sinθ=2/3
θ=41.8

b) Find the value of X

X=12cos180-θ
=12cos138.2
=-8.95??? Where did I go wrong?

For question use the below diagram,

I have marked an angle alpha on it.
Now resolve forces vertically to find angle alpha. Theta can be found by 180-alpha.

For b,
Resolve forces horizontally.
Substitute the value of alpha to find the answer.

Reply 2

Original post by najinaji

I'd like it if you could please check my answers & also provide help. :smile:

2)

Tension in AC=50N, Weight of particle C=W newtons

a) Find value of W

50-Wgcos60=0
Wgcos60=50
Wg=50/cos60=100
W=10.2

b) Find T of BC

T-WgSin60=0
T-100Sin60=0
T=86.6

For question 2, i will do the part b before a.
Resolve horizontally, you will get the tension in BC.

For part a,
Resolve vertically to find W.
NB: The weight is W. If it was said that the mass of C is m then you should have wrote the weight as mg, but here you will use W not Wg.

EDIT: I didn't went through your working before. You have done part (b) correctly. For part (a) your mistake is using Wg rather than W.

(edited 12 years ago)

Reply 3

najinaji

OP

18

Original post by raheem94

For question use the below diagram,

I have marked an angle alpha on it.
Now resolve forces vertically to find angle alpha. Theta can be found by 180-alpha.

For b,
Resolve forces horizontally.
Substitute the value of alpha to find the answer.

Thank you! :biggrin:

Original post by raheem94

For question 2, i will do the part b before a.
Resolve horizontally, you will get the tension in BC.

For part a,
Resolve vertically to find W.
NB: The weight is W. If it was said that the mass of C is m then you should have wrote the weight as mg, but here you will use W not Wg.

EDIT: I didn't went through your working before. You have done part (b) correctly. For part (a) your mistake is using Wg rather than W.

Ah, I see. So does that mean that W=100?

Reply 4

Original post by najinaji

Thank you! :biggrin:

Ah, I see. So does that mean that W=100?

Yes, W=100N.

Reply 5

Original post by najinaji

3)

Mass of parcel=5kg, tanα=3/4. Find the coefficient of friction between the parcel and the plane

F=20sinα, R=5gcosα
μ=F/R = 20sin36.87/5(9.8)cos36.87 = 0.306?

Can you please give the exact question. Your question doesn't states whether the particle is on the point of moving up the plane or moving down the plane.

Reply 6

Original post by najinaji

4)

tanθ=3/4

a) Show that the tension in the string is 5/2mg

b) Find the value of the coefficient of friction

I'm baffled by this question.

Your this question is also unclear.
Does the system contains 3 particles, A, B and C?

Reply 7

najinaji

OP

18

Original post by raheem94

Can you please give the exact question. Your question doesn't states whether the particle is on the point of moving up the plane or moving down the plane.

Sorry, it's on the point of sliding down.

Original post by raheem94

Your this question is also unclear.
Does the system contains 3 particles, A, B and C?

Yep. They're smooth rings & the coefficient of friction is mu between each ring & the pole.

Reply 8

Original post by najinaji

3)

Mass of parcel=5kg, tanα=3/4. Find the coefficient of friction between the parcel and the plane

F=20sinα, R=5gcosα
μ=F/R = 20sin36.87/5(9.8)cos36.87 = 0.306?

You should try to make your steps a bit more clear by indicating the direction in which you are resolving to get the equations.

Your both equations are wrong.

First resolve parallel to the plane, include friction, component of 20N force and component of weight. This will give you the friction.

Then resolve perpendicular to the plane, remember to include the normal reaction, component of weight and component of 20N force. This will give you the normal reaction.

Now use F = μR, you have both, F and R, so you can find μ.

Reply 9

Is 1) a, "θ" = 135/12 = 11.25

Am I incorrect?

Reply 10

Original post by makaveli33

Is 1) a, "θ" = 135/12 = 11.25

Am I incorrect?

How do you get 135/12?

Theta does not equal to 11.25degree, you are incorrect.

Reply 11

Original post by raheem94

How do you get 135/12?

Theta does not equal to 11.25degree, you are incorrect.

"a) Find the value of θ

12sinθ=8sin90
12sinθ=8
sinθ=2/3
θ=41.8"

I calculated Theta as.....
12sinTheta = 8sin 90
12sinTheta = 8
so....

135 = 8sin(90) + 4sin(45)
135 / 12sin = Theta
Theta = 11.25

However, this appears to be wrong, when I reiterate (relook at) it.

How about....

Is Theta...
27.86
I calculated this with....
41.8 / 0.6666...

But then when I look at above the first two lines of the question do not look right, ??? I am so confused.

Inclusive of this, the angle for Theta does not look like 27.86 at all =(

(edited 12 years ago)

Reply 12

Original post by makaveli33

"a) Find the value of θ

12sinθ=8sin90
12sinθ=8
sinθ=2/3
θ=41.8"

I calculated Theta as.....
12sinTheta = 8sin 90
12sinTheta = 8

Look at my first post in this forum, i have attached an image in it. In that image i have marked an angle alpha, where alpha = 180-theta. So if we find alpha then we can easily find theta.

Now try to do it by using alpha.

Hint: The angle theta is greater than 100degree.

Reply 13

najinaji

OP

18

Original post by raheem94

First resolve parallel to the plane, include friction, component of 20N force and component of weight. This will give you the friction.

Is it F-5gcosα+20cosα=0?

Reply 14

Original post by raheem94

Look at my first post in this forum, i have attached an image in it. In that image i have marked an angle alpha, where alpha = 180-theta. So if we find alpha then we can easily find theta.

Now try to do it by using alpha.

Hint: The angle theta is greater than 100degree.

Is theta, 135 degrees, by the way - excellent problem solving.

It could also possibly be 145 degrees = theta. As alpha looks like 35 or 45 (more likely 45 degrees), as it looks like 1/2 a right angle (alpha).

Thanks.

Reply 15

Original post by najinaji

Is it F-5gcosα+20~~cosα~~=0?

Look at the diagram below,

Notice that component of the 20N force that is perpendicular to the plane is 20sinα.

Reply 16

Original post by makaveli33

Is theta, 135 degrees, by the way - excellent problem solving.

It could also possibly be 145 degrees = theta. As alpha looks like 35 or 45 (more likely 45 degrees), as it looks like 1/2 a right angle (alpha).

Thanks.

In mechanics we don't make guesses, we try to resolve the forces and find the angles. You can only try to find the angles from the diagram if the question states that the diagram is to scale.

Reply 17