At the airport, you pull a 15kg suitcase across the floor with a strap that is at an angle of 45 degrees above the horizontal. Find the normal force and the tension in the strap if the suitcase is pulled at a constant speed and the coefficient of kinetic friction is 0.36.
Ws = M*g = 15 * 9.8 = 147 N. = Wt. of suitcase.
Fn = Ws-T*sin45 = 147 - T*sin45 = 147-0.707T = Normal force.
Fk=u*Fn=0.36*(147-0.707T) = 52.9-0.255T.
T*Cos45-Fk = M*a.
0.707T-(52.9-0.255T) = M*0.
0.707T-52.9+0.255T = 0.
0.961T = 52.9
T = 55 N.
Fn = 147 - 0.707*55 = 108 N.
You pull your 15 kg suitcase at the airport. What is the coefficient of kinetic friction (uk) between the case and the floor if you pull with a force of 50 N, and the suitcase accelerates at 0.52 m/s²
.52
Well, it seems like our suitcase is all set for its big flight! Let's do some math to find the normal force and the tension in the strap.
To start, we need to break down the forces acting on the suitcase. We have the force of gravity pulling it downwards, the normal force pushing it upwards, and the tension force pulling it forward.
Given that the angle between the strap and the horizontal is 45 degrees, we can break down the tension force into horizontal and vertical components. The horizontal component will help overcome the force of friction, while the vertical component will contribute to the normal force.
First, let's find the vertical component of the tension force. The total tension force can be found using the formula:
Tension force = (mass of the suitcase) × (acceleration due to gravity)
Tension force = (15 kg) × (9.8 m/s^2)
Tension force = 147 N
Since the angle of the strap with the horizontal is 45 degrees, the vertical component of the tension force will be:
Vertical component of tension force = Tension force × cos(45)
Vertical component of tension force = 147 N × cos(45)
Vertical component of tension force = 147 N × 0.707 (rounded)
Vertical component of tension force = 103.6 N (rounded)
This vertical component of the tension force is equal to the normal force acting on the suitcase. So, the normal force is approximately 103.6 Newtons.
Now, let's find the force of friction using the coefficient of kinetic friction. The force of friction can be calculated as:
Force of friction = (coefficient of kinetic friction) × (normal force)
Force of friction = 0.36 × 103.6 N
Force of friction ≈ 37.3 N
Since the suitcase is pulled at a constant speed, the force of friction must be equal to the horizontal component of the tension force. Hence, the horizontal component of the tension force is also approximately 37.3 Newtons.
In summary:
- The normal force acting on the suitcase is approximately 103.6 Newtons.
- The tension in the strap is approximately 37.3 Newtons (horizontal component).
Oh, and remember, always keep an eye on your luggage at the airport! Safe travels!
To find the normal force and tension in the strap, we can break down the forces acting on the suitcase.
First, we need to identify the forces involved. There are three forces acting on the suitcase: the force of gravity (mg), the tension in the strap (T), and the friction force (f).
Let's define the positive direction as the direction of motion (horizontal).
The force of gravity can be decomposed into two components: one parallel to the surface (mg sinθ) and one perpendicular to the surface (mg cosθ), where θ is the angle of the strap with the horizontal.
The tension in the strap is acting at an angle of 45 degrees above the horizontal, so its vertical component (T sinθ) opposes the force of gravity (mg cosθ), and its horizontal component (T cosθ) is in the same direction as the friction force.
Since the suitcase is moving at a constant speed, the sum of all the forces acting on it is zero. Therefore, we can write the following equations:
Sum of vertical forces = 0:
T sinθ - mg cosθ = 0 (Equation 1)
Sum of horizontal forces = 0:
T cosθ - f = 0 (Equation 2)
The friction force can be calculated using the equation:
f = μN
where μ is the coefficient of kinetic friction and N is the normal force.
Since the suitcase is not accelerating vertically, the normal force is equal in magnitude and opposite in direction to the vertical component of the force of gravity:
N = mg sinθ
Now, we can substitute the values into the equations to find the normal force and tension in the strap.
From Equation 2, we have:
T cosθ = f
Substituting the value of f from the friction equation, we get:
T cosθ = μN
Substituting the value of N, we get:
T cosθ = μ(mg sinθ)
Dividing both sides by cosθ, we get:
T = μmg tanθ
Now, substituting the given values:
T = (0.36)(15 kg)(9.8 m/s^2) tan(45°)
Calculating the value of T, we get:
T ≈ 75.03 N
To find the normal force, we use Equation 1:
T sinθ - mg cosθ = 0
Substituting the given values:
T sinθ - mg cosθ = 0
75.03 N sin(45°) - (15 kg)(9.8 m/s^2) cos(45°) = 0
Calculating the value of the normal force, we get:
N ≈ 109.51 N
Therefore, the normal force is approximately 109.51 N, and the tension in the strap is approximately 75.03 N.