Question

A water-skier is pulled behind a boat by a rope that is at an angle of 13° and has a tension of 490 N. The water-skier has a mass of 49 kg.

a. Draw a free-body diagram showing the forces acting on the skier. (She is being pulled to the right.)

b. What are the magnitudes of the x- and y-components of the tension?

c. What is the normal force acting on the skier?

Answer 1

try to draw an fbd with the water skier as a dot being pulled by a force at a 13 degree angle

Answer 2

the force is being applied at an angle

Answer 3

there are only two forces, the tension + gravity so there should only be two arrows

Answer 4

actually there's a normal force nvm 3 arrows

Answer 5

you don't have to draw the box every time, most fbs just have a dot

Answer 6

anyway, for b) to resolve the tension into its x/y components it's just Fcos(theta) and Fsin(theta) as usual

Answer 7

f would be 490 right?

Answer 8

yes

Answer 9

490sin(13)=110.23
490cos(13)=477.44

Answer 10

good, units are N as usual

Answer 11

to calculate the normal for c) we need to consider the y-direction of the tension force as well as gravity

Answer 12

net force = 0 = Fsin(theta) - mg + N

solve for N

Answer 13

110.23-49(9.8)+n

Answer 14

= 0 then solve for N

Answer 15

369.97

Answer 16

good so 369.97 N = your ans