Rachel pulls her 13 kg suitcase at a constant speed by pulling on a handle that makes an angle x with the horizontal. The frictional force on the suitcase is 28 N and Rachel exerts a 44 N force on the handle.

a) What angle does the handle make with the horizontal?

b) What is the normal force exerted on the suitcase?

Thoughts: normal force is equal to the weight minus the y component

(a) 44 cos x = 28 (horizontal equlibrium)

Solve for the angle x.
x = cos^-1 28/44 = 50.5 degrees

(b) use vertical equilibrium
up forces = down force

44 sin x + (normal force) = M g = 127.4 N

Solve for the normal force. You could also solve for the coefficient of kinetic friction, but they don't ask for that.

Thanks for your help.

To find the angle the handle makes with the horizontal, we can use the fact that the force Rachel exerts on the handle is equal to the horizontal component of the force equation.

a) The horizontal component of Rachel's force is given by Fhorizontal = F * cos(x), where F is the force Rachel exerts on the handle.

We are given that Fhorizontal = 44 N and we need to solve for x.
Using the cosine function, we can rearrange the equation to solve for x:

cos(x) = Fhorizontal / F
cos(x) = 44 N / 44 N
cos(x) = 1
x = cos^(-1)(1)
x = 0°

Therefore, the angle the handle makes with the horizontal is 0°.

b) The normal force exerted on the suitcase is equal to the weight of the suitcase minus the vertical component of the force equation.

The weight of the suitcase is given by W = m * g, where m is the mass of the suitcase and g is the acceleration due to gravity.

We are given that the mass of the suitcase is 13 kg. The acceleration due to gravity is approximately 9.8 m/s^2.

W = 13 kg * 9.8 m/s^2
W = 127.4 N

To find the vertical component of Rachel's force, we can use the sine function: Fvertical = F * sin(x), where F is the force Rachel exerts on the handle.

We are given that Fvertical = 44*sin(x).

Plugging in the angle x = 0°, the vertical component becomes:
Fvertical = 44 N * sin(0°)
Fvertical = 0 N

Therefore, the normal force exerted on the suitcase is equal to the weight of the suitcase (127.4 N) since there is no vertical force acting on it.

Hence, the normal force exerted on the suitcase is 127.4 N.

To solve this problem, we'll need to analyze the forces acting on the suitcase and use the given information to determine the angle and normal force.

a) To find the angle x, we need to consider the forces in the horizontal plane. The horizontal force on the suitcase is the sum of the force exerted by Rachel and the frictional force. Since the suitcase is moving at a constant speed, we know that the net force in the horizontal direction is zero. This means that the force exerted by Rachel should balance out the frictional force.

So, we can set up the following equation:

Force exerted by Rachel (44 N) + Frictional force (28 N) = 0

Solving for the force exerted by Rachel:

Force exerted by Rachel = - Frictional force
Force exerted by Rachel = - 28 N

Now we have the horizontal component of the force exerted by Rachel. To find the angle x, we can use trigonometry. The horizontal component of the force exerted by Rachel can be expressed as:

Horizontal component = Force exerted by Rachel * cos(x)

Substituting the known values:

-28 N = 44 N * cos(x)

Solving for cos(x):

cos(x) = -28 N / 44 N
cos(x) = -0.6364

Using inverse cosine (arccos) function on a calculator, we can find the angle x:

x = arccos(-0.6364)
x ≈ 132.5 degrees

Therefore, the angle the handle makes with the horizontal is approximately 132.5 degrees.

b) To find the normal force exerted on the suitcase, we need to consider the forces in the vertical plane. The suitcase is not moving vertically, so the net force in the vertical direction is zero. This means that the normal force should balance out the weight of the suitcase.

The weight of the suitcase can be calculated using the formula:

Weight = mass * gravity

Weight = 13 kg * 9.8 m/s^2
Weight ≈ 127.4 N

The normal force is equal and opposite to the weight, so the normal force exerted on the suitcase is approximately 127.4 N.

In summary:
a) The angle x that the handle makes with the horizontal is approximately 132.5 degrees.
b) The normal force exerted on the suitcase is approximately 127.4 N.