fisica cp 5

1. We are only concerned with horizontal forces in this problem (gravity plays no direct role). We take East as the +x direction and North as +y. This calculation is efficiently implemented on a vector-capable calculator, using magnitude-angle notation (with SI units understood). a= b

g b

g b

9.0 ∠ 0° + 8.0 ∠ 118°
F
=
= 2.9 ∠ 53° m 3.0

Therefore, the acceleration has a magnitude of 2.9 m/s2.

g

2. We apply Newton’s second law (specifically, Eq. 5-2).
(a) We find the x component of the force is
Fx = max = ma cos 20.0° = (1.00kg ) ( 2.00m/s 2 ) cos 20.0° = 1.88N.

(b) The y component of the force is
Fy = ma y = ma sin 20.0° = (1.0 kg ) ( 2.00 m/s 2 ) sin 20.0° = 0.684N.

(c) In unit-vector notation, the force vector (in newtons) is
.
F = Fx i + Fy j = 188i + 0.684 j .

3. We apply Newton’s second law (Eq. 5-1 or, equivalently, Eq. 5-2). The net force applied on the chopping block is Fnet = F1 + F2 , where the vector addition is done using

d

i

unit-vector notation. The acceleration of the block is given by a = F1 + F2 / m.
(a) In the first case
F1 + F2 =

i
( 3.0N ) ˆ + ( 4.0N ) ˆj

+

i
( −3.0N ) ˆ + ( −4.0N ) ˆj

=0

so a = 0 .
(b) In the second case, the acceleration a equals
F1 + F2
=
m

((3.0N ) ˆi + ( 4.0N ) ˆj) + (( −3.0N ) ˆi + ( 4.0N ) ˆj) = (4.0m/s )ˆj.
2

2.0kg

(c) In this final situation, a is
F1 + F2
=
m

( (3.0N ) ˆi + ( 4.0N ) ˆj) + ((3.0N ) ˆi + ( −4.0N ) ˆj) = (3.0 m/s )i.
ˆ
2

2.0 kg

4. The net force applied on the chopping block is Fnet = F1 + F2 + F3 , where the vector addition is done using unit-vector notation. The acceleration of the block is given by a = F1 + F2 + F3 / m.

d

i

(a) The forces (in newtons) exerted by the three astronauts can be expressed in unitvector notation as follows:

(
)
F = 55 ( cos 0°ˆ + sin 0°ˆ ) = 55 ˆ i i j F = 41 ( cos ( −60° ) ˆ + sin ( −60° ) ˆ ) = 20.5 ˆ − 35.5 ˆ i i
j.
j
F1 = 32 cos 30°ˆ + sin 30°ˆ =