Hey, so here here are the solutions:
13F q14: This seriously needs a diagram. Using it, you can work out all the angles needed, and then resolve the forces into horizontal/vertical components.
Let T be the tension in the string, and P be the force used to pull the particle.
P is applied @ 75 degrees to the vertical, so it is @ 15 degrees to the horizontal
To find the angles for the T triangle, use
 = \frac{2}{2.5})
and
Now resolve vertically and horizontally:
Vertically:
-2g-P\sing(15)=0)
Horizontally:
=0)
Solve simultaneously [cbf typing this out on LATEX], and you get:
and
13F q15 You'll find that the particle rests in the shape of a pythagorean triangle, with side 13cm, 5cm, and 12cm
Find the angles (67.38 and 22.68 degrees {i.e. use tan = 12/5}) and the resolve components:
Let the TEnsion in the 5cm string = A, and the tension in the 12cm = B
Horiz:
-B\cos(22.68)=0)
Vert:
+B\sin(22.68)=5g)
Solve simultaneously to get A = 45.23N and B = 12.18N
13G 3c - okay, so the way this differs from 3b, is that Pushing force now how a downward component that affects the magnitude of the reaction/normal force
REsolve P into horiz and vert components: vert = Psin60 and horiz = Pcos60, and now set up 2 equations:
Vertical forces:
=100g)
...(1)
Horizontal:
=0.3N)
... (2)
therefore,
, and sub this into (1)
...(1)
and P = 1224N
13G 4 The frictional force simply equals the weight component of the particle parallel to the plane (because net force = 0)
so, W(parallel) = 5sin(30) = 2.5N = Fr
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