Mechanics Question Help

Firefox23

Could someone help me out with the second part of this question?

(edited 8 years ago)

Screen Shot 2015-09-16 at 17.41.48.png39.1KB

Scroll to see replies

Reply 1

TeeEm

Original post by Firefox23

Could someone help me out with the second part of this question?

you need to attach it...

Reply 2

Zacken

What ave you tried so far?

Reply 3

Firefox23

Original post by Zacken

What ave you tried so far?

I initially tried to find the acceleration using F=man with F as the weight down the plane - the friction up the plane and used s=ut +1/2at^2 with s as 5a and u as 0

Reply 4

雷尼克

draw a clear diagram first then label all of the forces along each plane and you should be able to do this rather easily.

Reply 5

Zacken

Original post by Firefox23

I initially tried to find the acceleration using F=man with F as the weight down the plane - the friction up the plane and used s=ut +1/2at^2 with s as 5a and u as 0

Okay, so we're taking down the slop as positive then - this means we have

s = 5a, u=0, a = a, t=T

so we need only find acceleration then.

Let's see - we resolve down the plane to get

mg \sin \theta - \mu R

where we can get R easily, simply resolve perpendicular to the plane to get that

R = mg \cos \theta

So our force down the plane is

Unparseable latex formula:

mg \sin \theta - \frac{1}{4} \mg \cos \theta

where

\theta = \arctan \frac{3}{4}

.

Hence you get (after dividing by m)

a = \frac{2g}{5}

.

Using the suvat equation yields

5a = \frac{1}{2} \cdot \frac{2g}{5} \cdot T^2 \implies T = 5\sqrt{\frac{a}{g}}

That's the time down the slope, they are asking for the total time - so you need to add the time to go up the slope as well.

That's a basic suvat solve for let's call it

t_{up}

, which you should get as

Spoiler

Then you add and pull out a common factor.

Reply 6

Firefox23

Original post by Zacken

Okay, so we're taking down the slop as positive then - this means we have

s = 5a, u=0, a = a, t=T

so we need only find acceleration then.

Let's see - we resolve down the plane to get

mg \sin \theta - \mu R

where we can get R easily, simply resolve perpendicular to the plane to get that

R = mg \cos \theta

So our force down the plane is

Unparseable latex formula:

mg \sin \theta - \frac{1}{4} \mg \cos \theta

where

\theta = \arctan \frac{3}{4}

.

Hence you get (after dividing by m)

a = \frac{2g}{5}

.

Using the suvat equation yields

5a = \frac{1}{2} \cdot \frac{2g}{5} \cdot T^2 \implies T = 5\sqrt{\frac{a}{g}}

That's the time down the slope, they are asking for the total time - so you need to add the time to go up the slope as well.

That's a basic suvat solve for let's call it

t_{up}

, which you should get as

Spoiler

Then you add and pull out a common factor.

Thanks so much I forgot to add the time up the slope to get the total time. Any chance you could help me with this question? I'm not sure where to start.

FullSizeRender-10.jpg51.3KB

Reply 7

ghostwalker

Original post by Firefox23

Any chance you could help me with this question? I'm not sure where to start.

Start at A.

Since you have symmetry, what's the vertical reaction at A?

As A and E are as far apart as possible what's the frictional force going to be at A?

What's the vertical component in AB?

You now have sufficient information to work out the angle of AB, and hence the horizontal displacment of B.

Now go on to consider B and BC.
PS Not worked that bit through.

(edited 8 years ago)

Reply 8

Firefox23

Original post by ghostwalker

The tension in each section of the string will be different right? Also what do you mean by the vertical component in AB? This is what I tried . . .

IMG_6924.jpg510.7KB

Reply 9

ghostwalker

Original post by Firefox23

The tension in each section of the string will be different right?

The tension in the upper strings will be different to that in the lower strings.

Also what do you mean by the vertical component in AB? This is what I tried . . .

The tension in AB can be split into vertical and horizontal components.

You can write down the vertical reaction at A straight off. What's the total weight? Then by symmetry vertical reaction at A is ...?

Then as per my previous questions.

Note: There is no

T_1

force, only friction and the horizontal component in AB.

Reply 10

Firefox23

Original post by ghostwalker

The tension in the upper strings will be different to that in the lower strings.

The tension in AB can be split into vertical and horizontal components.

You can write down the vertical reaction at A straight off. What's the total weight? Then by symmetry vertical reaction at A is ...?

Then as per my previous questions.

Note: There is no

T_1

force, only friction and the horizontal component in AB.

This is what I've tried but I still seem to have too many unknowns . . .

Reply 11

ghostwalker

Original post by Firefox23

This is what I've tried

Doesn't seem that way from your working - unless you've done something since my last post, but haven't posted it.

Reply 12

Firefox23

Original post by ghostwalker

Doesn't seem that way from your working - unless you've done something since my last post, but haven't posted it.

So my working's correct?

Reply 13

ghostwalker

Original post by Firefox23

So my working's correct?

Sorry, didn't see your new working at the top of the post - will have a look now.

Reply 14

ghostwalker

Original post by Firefox23

So my working's correct?

You have ignored what I suggested.

The total weight of the mass is 8mg.
The vertical reactions at A and E must add to 8mg.

By symmetry the vertical reaction at A will be half that, i.e. 4mg.

Reply 15

Firefox23

Original post by ghostwalker

You have ignored what I suggested.

The total weight of the mass is 8mg.
The vertical reactions at A and E must add to 8mg.

By symmetry the vertical reaction at A will be half that, i.e. 4mg.

Does the tension in AB not affect the vertical reaction? Sorry I didn't misread the question and didn't notice that the A and E had different masses to B,C and D

Reply 16

ghostwalker

Original post by Firefox23

Does the tension in AB not affect the vertical reaction? Sorry I didn't misread the question and didn't notice that the A and E had different masses to B,C and D

The tension in AB does effect the vertical reaction. The vertical reaction at A, 4mg, arises as a result of the weight of A, mg, and the vertical component of the tension in AB.

Reply 17

Firefox23

Original post by ghostwalker

The tension in AB does effect the vertical reaction. The vertical reaction at A, 4mg, arises as a result of the weight of A, mg, and the vertical component of the tension in AB.

I think I understand R=4mg and it also = mg +tcostheta

Reply 18

ghostwalker

Original post by Firefox23

I think I understand R=4mg and it also = mg +tcostheta

Yep.

In the first case you look at the structure as a whole, and in the second you just focus on the details at A.

Reply 19

Firefox23

Original post by ghostwalker

Yep.

In the first case you look at the structure as a whole, and in the second you just focus on the details at A.

Thanks I've worked it through now, any chance you could help me out with this question if you have time? I've tried to do velocity triangles and I've tried i and j notation but I can't seem to get the right answer. Its question 14.