Sample Test answers:

1) First find the components of F₁ and F₂, then F₃ = -(F₁ + F₂) because the sum of all forces (the net force) is zero: F₁ + F₂ + F₃ = 0.
F₃ = (-33.5 - 8.8) N (or 34.6 N at 14.7° below the -x axis)
2) The dog’s weight is 33 lbs. The person’s mass is 65.3 kg.

3) The maximum frictional force is = 9.8N. Once the block is sliding, =1.96 N

4) See the book!

5) Set up Newton’s second law for each block:

block 1: F_T = m₁a
block 2: F_T - m₂g = -m₂a
Then solve for a: a = m₂g/(m₁ + m₂) = 3.27 m/s²
Now get the tension from the first equation: F_T = 32.7 N

6) Find the acceleration from v_f = v_i + a t, (a = 5 m/s²), then F = ma = 3500 N

7) For each case, use Newton’s 2nd law: (F_N - m g = m a) (the scale measures the normal force, not the weight, or you could call it the “apparent weight”)
At rest or at constant velocity, a=0, so, the scale reads (F_N = m g = 784 N) for (a), (b), and (c)

For (d) F_N = 904 N, (e) F_N = 744 N, (f) F_N = 0 N