Physics 101 - Homework # 3 Solutions

• Problem 4-4
• Problem 4-15
• Problem 4-17
• Problem 4-21
• Problem 4-22
• Problem 4-24
• Problem 4-25
• Problem 4-43
• Problem 4-51
• Problem 4-56

Chapter 4

Problem 4)

Since the velocity is constant, F = 0 (no acceleration); the force from the water must therefore be equal in magnitude and opposite in direction to the wind force, i.e. it must be 10⁴ N from the west.

Problem 15)

Problem 17)

The tension in the pulled rope must be equal to the force the father exerts: T = F, as long as the rope is massless.

We need to consider the horizontal forces only (there will be no motion in the vertical direction). If we take both sleds as the object, we get the first force diagram shown below.

Then for the horizontal motion, we have

F = T cos30° = (m + m)a = 2ma
so    a = T(cos30°)/2 m = 0.433T/m

If we take the second sled as the object, we get the second force diagram shown above.

Then

F = T₂ = ma

so

T₂ = m(0.433 T/m) = 0.433 F

Problem 21)

The force the automobile exerts on the earth is the reaction to the force of gravity on the automobile (Newton's third law), so the two forces must have the same magnitude:

F_auto = F_earth

m_autoa_auto = m_eartha_earth

(950 kg)(9.8 m/s²) = (6.0 x 10²⁴ kg)a_earth

which gives    a_earth = 1.6 x 10^-21 m/s²

No wonder it is not noticeable!

Problem 22)

The force exerted by the satellite on the astronaut is the reaction to the force of the astronaut on the satellite (3rd law again...):

F_astronaut = - F_satellite = (-0.5i - 0.7j - 5.1k) N

with magnitude    5.17 N    [ magnitude = {(-0.5² + (-0.7)² + (-5.1)²} ^1/2 ]

Problem 24)

Since we need only the horizontal forces (the vertical normal forces balance the gravity forces), diagrams for the set and each of the blocks are as shown.

a) For the set of blocks we have (where M= total mass)

F_x = M a_x

8.0 N = (2.0 kg + 3.0 kg + 4.0 kg)a

which gives    a = 0.89 m/s²

b) For block 1 we have F_x = m a_x

F - F₁₂ = m₁ a    (where F₁₂ is the contact force exerted on block 1 by block 2)

8.0 N - F₁₂ = (2.0 kg)(0.89 m/s²)

which yields

F₁₂ = 6.2 N   to the left

The forces are F = 8.0 N to the right, F₁₂ = 6.2 N to the left, and the net force is F_{1 net} = 1.8 N to the right.

c) For block 2 we have

F_x = ma_x

F₂₁ - F₂₃ = m₂a     and F₂₁ = F₁₂    (Newton's 3rd law)

6.2 N - F₂₃ = (3.0 kg)(0.89 m/s²)

which gives F₂₃ = 3.5 N to the left

The forces are F₂₁ = 6.2 N to the right, F₂₃ = 3.5 N to the left, and the net force is F_{2 net} = 2.7 N to the right.

d) For block 3 we have F_x = ma_x    so   F₃₂ = m₃a

F₃₂ = (4.0 kg)(0.89 m/s²)

which gives

F₃₂ = 3.6 N to the right ( = F₂₃, Newton's 3rd law)

The forces are F₃₂ = 3.6 N    to the right and the net force is F_{3 net} = 3.6 N to the right.

Problem 25)

a) If we look at the five cars, the only horizontal force is the force between the engine and the first car. We have

F_x = F = m_carsa_x

4.5 x 10⁴ N = 5(3.0 x 10⁴ kg)a_x

which yields

a_x = 0.30 m/s²

This is the acceleration of the engine and of each of the cars (they are coupled together, so they must all have the same acceleration).

b) If we look at the first car, the horizontal forces are F_engine caused by the engine, acting to the right, and F_cars the force exerted to the left by the other four cars. Applying F=ma to the first car, we have the horizontal-component equation:

F_engine - F_cars = m₁a

4.5 x 10⁴ N - F_cars = (3.0 x 10⁴ kg)(0.30 m/s²)

which gives

F_cars = 3.6 x 10⁴ N   backward

Problem 43)

See Diagram:

If the ground is smooth, the horizontal frictional force f is very small; if this becomes less than the normal force exerted by the wall, then there will be a non-zero net horizontal force, and thus an acceleration in the x-direction.

Problem 51)

We can find the acceleration of the object by differentiating the displacement function:

v = dx/dt = d(At^3/2)/dt = (3/2)At^1/2

a = dv/dt = d((3/2) At^1/2)/dt = (3/4)At^-1/2

We find the net force from

F_net = ma

= m(3/4)At^-1/2

= (3/4)(2.0 kg)(6.0 m/s^3/2)t^-1/2

= (9.0 Ns^1/2)t^-1/2

(note that this is a non-constant force)

Problem 56)

From the force diagram for the pulley and object we can write

F_x = T₂cos35° - T₁cos35° = 0    which gives T₁ = T₂

F_y = T₁sin35° + T₂sin35° - F_g

= 2T₂ sin35° - (20 N)

which gives T₂ = 17.4 N

From the force diagram for the other object we can write

F_y = T₂ - F_g' = 0

which gives

F_g' = T₂ = 17.4 N