How will you solve these?

Question

1. marc and john pull at two ropes attached to a box of fruits with forces of 60 n and 100 N RESPECTIVELY. find the resultant if they pull a.) in the same direction b.) in the opposite directions c.) at right angle to each other and at an angle of 60 degrees between them.

2. janet walks 6km east then 3km south and finally 6 km west. find her final displacement .

3. minnie dives of the 4m springboard and initially bounces up with a velocity of 8m/s at an angle of 60 degrees to the horizontal. what are the horizontal and vertical components of her velocity?

Answer 1

1

a

160 N

b

40 N

c

can't be at right angles to each other and 60 degrees at the same time

2

3km

3

horizontal -> 8cos60 = 4m/s

verticle -> 8sin60 = 6.9282m/s

Answer 2

This sounds like an exercise is adding vectors.  Basically the idea is to add both the magnitudes and directions together separately.

If both Marc and John pull together in the same direction, you add directly:

a) 60 + 100 = 160 N in the direction they are pulling

Since they are now pulling opposite, the resultant is the difference between the two:

b) 100 - 60 = 40 N in the direction of John

c)  For this question, the two vectors need to be added, but it is not quite as straight forward.  The overall force will not be a direct addition of the values, but rather the "hypotenuse" as they are at right angles to each other.

F = sqrt(100^2 + 60^2) = 116.6 N

To find the angle, we can use trigonometry knowing that tan(a)=F1/F2.

a = atan(100/60) = 59.0 degrees from Marc.  This would be 31 degrees from John.

d)  To solve this, we can use the law of cosines:

C^2=A^2+B^2-2ABcos(c)

C^2 = 100^2+60^2-2(100)(60)cos(180-60)

C = 140 N

The direction can be found using the law of sines:

C/sin(c) = A/sin(a)

140/sin(180-60) = 100/sin(a)

sin(a) = 0.6186

a = 38.2 degrees from Marc

Note that this method could have also been used for the two boys pulling at right angles, the -2ABcos(c) term just equates to 0.

2.  If Janet walks 6 km east, then 3 south, then 6 km west she has a resultant displacement of 3 km south.  The 6 km east and 6 km west will directly cancel out with each other, only leaving the 3 km south.

3. The initial velocity is 8m/s, which is the vector sum of the horizontal and vertical velocity components.  Once again, we will have to use trig.

Vy = 8sin(60) and Vx = 8cos(60)

Vy = 6.93 m/s upwards

Vx = 4.00 m/s horizontally

Note that this method could have been applied to part 1d, we could have taken both forces and broken them down into their x and y components.  Then you would add the two x components together and then add the two y componenets together.  Then we could use the hypotenuse method of part c to calculate the resultant.  This would avoid using the "tricky" law of cosines/sines.