# M1 -> Forces/Motion in 2D

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#1
Hi!

Struggling with these M1 Q's, could someone help? Appreciate it if poss Quite chunky Q's so excuse the mess. Apologies for poor pics

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The tension in the string is T1, where T1 = 30i + 49k. The horizontal force is F1, where F1 = pi.
1. i) Write down the value of p and show that the mass of the box is 5kg.
1. ii) Calculate the magnitude of T1 and the angle that T1 makes with the horizontal.

Another force, F2 = 48i - 87k is now applied to the box. The force F1 still acts and the box is still in equilibruim.

1. iii) The new tension in the string is T2 = ai + bk. Calculate the values of a and b.

The tension in the string now becomes T3 = q(13i + 84k), where q is a positive constant. The magnitude of this tension is 340N. The forces F1 and F2 remain unaltered.

1. iv) Find the value of q. Find also the acceleration of the box in ms-2, giving your answer in the form ci + dk, where c and d are to be deteremined.

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The diagram shows a load of weight WN at B in equilibrium. The load is attached by one light string to the ceiling at A and by a second light string that passes over a smooth pulley at C to an object of mass MKG hanging freely at D. The angles of the strings and the tension T1N and 196N acting at B are shown in the diagram.

2. i) Write down the numerical value of the tension in the string section CD, giving a reason for your answer. By considering the equilibrium of the object at D, calculate the value of m.

2. ii) Calculate the value of T1.

2. iii) Calculate the value of W.

2. iv) Calculate the magnitued of the total force exerted on the pulley at C by the string passing over it.

2. v) An addition mass, MKG, is now added at D. Explain why the system cannot be in equilibrium with ABC in a straight line no matter what the value of M.

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That's all. I'm so confused Could someone help me?

0
14 years ago
#2
I ve never really seen m1 questions like these , what board are you doing?

For first question if T1 = 30i + 49k then tan x= 49/30 , you should then be able to resolve horizontally and vertically(assuming its in equilibriam) for W( t1 sin x =W)and F1(t1 cos x=f1).
0
14 years ago
#3
b) x= arctan (49/30)

magnitude is sqrt(30^2 +49^2)

next bit

F2 = 48i - 87k

arctan to get angle then resolve to find horizontal component with cos x
use this horizontal component (may have to change it's direction ) to get T by trig.
0
#4
Evening!

I've managed to do all of #1 now thank you for the help.

Could anyone please help with question #2? Just some hints and pushes in the right direction would be great

0
14 years ago
#5
A good first step would be to resolve all the forces you have. Then Newton's 2nd and 3rd laws will help.

Also, you haven't really made it clear where T_1 acts, but I assume it acts in AB, otherwise we couldn't really just "write down" the tension in CD. (This could serve as another hint. )
0
#6
Evening!

<edited out>

This question is rather confusing in all honesty!

I'll go off and resolve them anyways, any more helps for the other parts would be great!

-edit-

Ok i'm confused
0
14 years ago
#7
Note that you can consider the particle at D and the string to separately, so that mg = 196. So this gives you m.

Now use the fact that forces up = forces down, and forces to the left = forces to the right, since the system is in equilibrium. This should get you T_1 and W.

And the force exerted on the pulley is just the reaction force due to the tension in the string BC and the weight of the particle hanging.

Finally, for the last part, the system changes. So draw the line and use Newton's laws again.
0
2 years ago
#8
How did you do 1) iv) Please explain.
0
7 months ago
#9
Please could u explain ur hint to the last part I do not understand what you mean.Thank you
0
7 months ago
#10
Can I get help with the iv and v.

Thank you
0
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