- #1

- 10

- 0

## Homework Statement

In chapter 5, problems 26:

A provided figure shows an overhead view of a 0.0250 kg lemon half and two of the three horizontal forces that act on it as it is on a frictionless table. Force F_1 has a magnitude of 6.00 N and is at θ_1 = 30.0° (and, btw, is in quadrant II). Force F_2 has a magnitude of 7.00 N and is at θ_2 = 30.0° (in quadrant IV). In unit-vector notation, what is the third force if the lemon half (a) is stationary, (b) has constant velocity v= (13.0 i- 14.0 j) m/s, and (c) has varying velocity v= (13.0t i- 14.0t j) m/s^2, where t is time?

## Homework Equations

F_net = m*a

F_net = F_1 + F_2 + F_3

F_3 = ?

F_3 = F_net- F_1 - F_2

needed: F_3x = m*a_x - F_1*cosθ_1 - F_2sinθ_2

F_3y = m*a_y - F_1*sinθ_1 - F_2cosθ_2

## The Attempt at a Solution

(a) acceleration = 0, therefore:

F_3 = 9.00 N i- 9.06 N j

(b) acceleration is still zero. so i assume part (a) = part (b)

(c)

F_net= m*a

m = 0.0250 kg

a = (sqrt) [13.0^2 + (-14.0)^2] = 19.1 m/s^2

F_net=0.4776 N

F_3= -8.37 N i- 9.41 N j

## Homework Statement

(Second problem) Chapter 5, problem 36:

An elevator cab is pulled upward by a cable. The cab and its single occupant have a combined mass of 2000 kg. When that occupant drops a coin, its acceleration relative to the cab is 8.00 m/s^2 downward. What is the tension in the cable?

## Homework Equations

I used:

T= mg-ma

## The Attempt at a Solution

T= 2000 kg (9.80 m/s^2)- 2000 kg (8.00 m/s^2)

= 3600 N

---please HELP!:surprised