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Physics: Post your doubts here!

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sagar65265

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amal sharkawi said:
View attachment 45105

can any one answer this plz??
Click to expand...
forces-on-ball-jpg.44836


Since the ball is balanced and in equilibrium, we can say that the net force on the ball is equal to zero. We can also split this net force into two components, one in the up-down (vertical) direction and another in the left-right (horizontal) direction. Let's also say that the upwards direction is positive and the leftward direction is negative.

In the vertical direction:

i) T has a component of T * cos(30) = (√3/2) * T = √3T/2 = (+√3T/2) Newtons
ii) W has a component of 0.15 Newtons downwards, so it has a component of (-0.15) Newtons.
iii)The wind force is horizontal, so it has no component.

Summing these up should give us zero, so

(+√3T/2) + (-0.15) = 0
√3T/2 = 0.15
√3T = 0.3

T = 0.1732 Newtons

That is the magnitude of the Tension force. Now let's do the horizontal direction:

i) T has a component of T * sin(30) in the rightward direction = T * (1/2) = T/2. Since this is to the right, the sign is negative, i.e. (-T/2) Newtons.
ii) The weight has no horizontal component in this situation.
iii) The Wind force has a component F(air) to the left, and only the left. Therefore, it has a component of (+F(air)) Newtons.

Summing these up should give zero, so

(+F(air)) + (-T/2) = 0
F(air) = T/2 = 0.1732/2 = 0.866 Newtons = 0.87 Newtons = B.

Hope this helped!
Good Luck for all your exams!
 
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sagar65265

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amal sharkawi said:
View attachment 45111


any one answer this plz
Click to expand...

23)

When the force W is applied on R, the spring R gets the full load of the force. In other words, the force extending R is equal to the force W. Therefore, the extension of spring R is equal to (using the formula |F| = kx):

x = |F|/k = W/k = W/3k (since the spring constant of R is 3k).

Now, the spring R will exert the same force at both ends - it will exert a force W on the lower end, and it will exert a force W on the upper end, i.e. on the bar.
For the bar to be in equilibrium, the force on it have to balance out, so

(Force from spring P) + (Force from spring Q) = (Force from spring R) (The bar has negligible weight)

For moments to be zero about the center of mass, the (Force from spring P) has to be equal to the (Force from spring Q). Therefore,

2(Force from spring P) = 2(Force from spring Q) = W

Therefore, the (Force from spring P) = W/2 = (Force from spring Q).

The extension of spring P if it has to exert a force of W/2 is equal to

x = |F|/k = (W/2)/k = W/2k

The extension of spring Q is the same, so that the system is stable and balanced.

Therefore, the overall extension is

W/2k + W/3k = 3W/6k + 2W/6k = 5W/6k meters = A.
If you still have a doubt concerning either of these questions, just post it on the forums and i'll see if I can answer.

Hope this helped!
Good Luck for all your exams!
 
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sagar65265 said:
23)

When the force W is applied on R, the spring R gets the full load of the force. In other words, the force extending R is equal to the force W. Therefore, the extension of spring R is equal to (using the formula |F| = kx):

x = |F|/k = W/k = W/3k (since the spring constant of R is 3k).

Now, the spring R will exert the same force at both ends - it will exert a force W on the lower end, and it will exert a force W on the upper end, i.e. on the bar.
For the bar to be in equilibrium, the force on it have to balance out, so

(Force from spring P) + (Force from spring Q) = (Force from spring R) (The bar has negligible weight)

For moments to be zero about the center of mass, the (Force from spring P) has to be equal to the (Force from spring Q). Therefore,

2(Force from spring P) = 2(Force from spring Q) = W

Therefore, the (Force from spring P) = W/2 = (Force from spring Q).

The extension of spring P if it has to exert a force of W/2 is equal to

x = |F|/k = (W/2)/k = W/2k

The extension of spring Q is the same, so that the system is stable and balanced.

Therefore, the overall extension is

W/2k + W/3k = 3W/6k + 2W/6k = 5W/6k meters = A.
If you still have a doubt concerning either of these questions, just post it on the forums and i'll see if I can answer.

Hope this helped!
Good Luck for all your exams!
Click to expand...
Sagar!
Check is this method correct... o_O or I did it wrong :'(

-------------------------------------------------------------------------
We use F = kx
so we know k for P and Q is k and for R is 3k
--> Therefore total extension of P and Q = k + k = 2k and extension for P,Q and R = 1/2k + 1/3k = (5/6)^-1k = 6/5k
so F = W, k = 6/5k and x = ?
W = 6/5k * x
x = 5w/6k
 
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sagar65265

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Thought blocker said:
Sagar!
Check is this method correct... o_O or I did it wrong :'(

-------------------------------------------------------------------------
We use F = kx
so we know k for P and Q is k and for R is 3k
--> Therefore total extension of P and Q = k + k = 2k and extension for P,Q and R = 1/2k + 1/3k = (5/6)^-1k = 6/5k
so F = W, k = 6/5k and x = ?
W = 6/5k * x
x = 5w/6k
Click to expand...

Yep! (y) That's alright, no probs!

Good Luck for all your exams!
 
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