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Question

1)A 10 Kg block is placed on a floor (co efficient of friction between the block and floor is 0.25). It is pulled horizontally by a force of 2 kg-wt. What is the frictional force acting at the interface?

2)Can you add 3 unit vectors to get a unit vector?

3)Two vectors 2p and p are inclined to each other at a certain angle. If the first vector be doubled, then the resultant vector is increased three times. Calculate the angle between the two vectors.

Answer 1

1) The frictional force is equal to umg, where u is the coefficient of friction, m is mass, and g is gravity. In this case, 2 kg-wt is already mg. umg = 0.25*10 = 2.5 kg-wt. That is greater than the applied force, so the block will not move. Normally, to solve this problem, you would need the coefficients of static and kinetic friction, respectively. You would first calculate if the applied force can overcome static friction. If not, static friction is the frictional force. If so, you need to calculate the force of kinetic friction instead. But if the two coefficients are equal, you don't need to go through that.

2) Yes. You can put the first two unit vectors in any direction you want, and then place the third one running exactly opposite the second one, getting you back to the end point of the first unit vector. The net result is therefore a unit vector.

3) I'm assuming the problem means that the magnitude of the resultant vector is tripled; its direction will definitely change. The length of the resultant can be found using the rule of cosines. I'll us p to represent the length of vector p. c^2 = a^2 + b^2 - 2ab*cos(C). In this case, we have c^2 = p^2 + (2p)^2 - 2p(2p)*cos(C) and (3c)^2 = (2p)^2 + (2p)^2 - 2(2p)(2p)*cos(C). First, I'll simply these expressions a little bit. c^2 = p^2 + 4p^2 - (4p^2)*cos(C) = 5p^2 - (4p^2)*cos(C) and 9c^2 = 4p^2 + 4p^2 - (8p^2)*cos(C) = 8p^2 - (8p^2)*cos(C). I can combine these into a single expression, 8p^2 - (8p^2)*cos(C) = 9[5p^2 - (4p^2)*cos(C)] ==> 8p^2 - (8p^2)*cos(C) = 45p^2 - (36p^2)*cos(C) ==> (28p^2)*cos(C) = 37p^2 ==> cos(C) = 37/28. So C = arccos(37/28). You can use a calculator to caclucate this value in degrees or radians.

Answer 2

The frictional force depends only on the normal force which is 10*g newtons, so the frictional force is 0.25*10*g newtons.

The magnitude of the sum of the three unit vectors (x,y,z) is √[1^2+1^2+1^2] = √3

The vector sum is the vector joining the base of the first to th tip of the second if the vectors are placed end-to-end. This forms a triangle, with the angle opposite the sum vector = 180º- ø, where ø is the angle between the vectors. By the law of cosines, the length of that sum is

L1^2 = (2p)^2 + p^2 - 2*(2p)*p*cos(180-ø)
L2^2 = (4p)^2 + p^2 - 2*(4p)*p*cos(180-ø)

Note that cos(180º-ø) = -cosø

L1^2 = 4*p^2 + p^2 + 4*p^2*cosø
l2^2 = 16*p^2 + p^2 + 8*p*cosø

L2 = 3*L1, L2^2 = 9*L1^2, so

9*L1^2 = 36*p^2 + 9*p^2 + 36*p^2*cos(ø)]

45*p^2 + 36*p^2*cosø =
17*p^2 + 8*p*cosø

28*cosø = 17 - 45 = -28

cosø = -1, ø = 180º

This of course should be obvious from examining the geometry of two colinear vectors. If the vectors are in line, the sum is simply the algebraic sum of their magnitudes. ø = 180º means the second vector is opposite in sign to the first. In the initial case, the sum is 2p - p = p. Now double the first to get 4p - p = 3p; doubling the first tripled the sum.

Answer 3

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