At the airport, you pull a 18 kg suitcase across the floor with a strap that is at an angle of 41° above the horizontal. Find the normal force and the tension in the strap, given that the suitcase moves with constant speed and that the coefficient of kinetic friction between the suitcase and the floor is 0.31. normal force N
tension N
Well, well, well, look who's got some physics questions! Don't worry, I've got just the right equation for this situation. Let's break it down, shall we?
First things first, we need to find the force of gravity acting on the suitcase. Since the suitcase has a mass of 18 kg, we can calculate its weight using the formula weight = mass × gravitational acceleration. The gravitational acceleration is approximately 9.8 m/s² on Earth.
So, weight = 18 kg × 9.8 m/s² = 176.4 N
Now, let's tackle the normal force. The normal force is the force exerted by a surface that is perpendicular (normal) to that surface. In this case, it's the floor.
Since the suitcase is being pulled at an angle of 41° above the horizontal, we need to find the vertical component of the tension in the strap. The vertical component of the tension is equal to the normal force.
Using trigonometry, we can find that the vertical component of the tension is given by the equation tension × sin(41°).
Since the suitcase is moving at a constant speed, the force of kinetic friction must be equal to the force of tension. So we can equate the force of kinetic friction (µk × normal force) to the tension × sin(41°).
Given that the coefficient of kinetic friction between the suitcase and the floor is 0.31, we can substitute this value into the equation:
0.31 × normal force = tension × sin(41°)
Now, we have two equations: weight = normal force and 0.31 × normal force = tension × sin(41°).
By substituting weight into the second equation, we can solve for the tension:
0.31 × weight = tension × sin(41°)
0.31 × 176.4 N = tension × sin(41°)
54.624 N = tension × sin(41°)
To find the normal force, we substitute the weight into the first equation:
normal force = 176.4 N
So, the normal force is 176.4 N and the tension in the strap is approximately 54.624 N. Voila! Done and dusted.
To find the normal force and the tension in the strap, we can use the following steps:
Step 1: Resolve the weight of the suitcase into its horizontal and vertical components. The vertical component, which is the weight acting downwards, is given by the formula:
Weight (vertical component) = mass * acceleration due to gravity = 18 kg * 9.8 m/s^2 = 176.4 N
Step 2: Determine the horizontal component of the weight. The horizontal component is given by the formula:
Weight (horizontal component) = Weight * cos(angle)
Weight (horizontal component) = 176.4 N * cos(41°)
Weight (horizontal component) = 176.4 N * 0.753
Weight (horizontal component) = 132.79 N
Step 3: Calculate the frictional force acting on the suitcase. The frictional force is given by the formula:
Frictional force = coefficient of friction * Normal force
Since the suitcase is moving at a constant speed, the frictional force is equal to the horizontal component of the weight:
Frictional force = 132.79 N
Step 4: Determine the normal force. The normal force is the force exerted by the surface of the floor on the suitcase and is equal in magnitude and opposite in direction to the vertical component of the weight. Therefore:
Normal force = Weight (vertical component) = 176.4 N
Step 5: Find the tension in the strap. The tension in the strap is the sum of the horizontal component of the weight and the frictional force:
Tension in the strap = Weight (horizontal component) + Frictional force
Tension in the strap = 132.79 N + 132.79 N = 265.58 N
Therefore, the normal force is 176.4 N and the tension in the strap is 265.58 N.
To find the normal force and tension in the strap, we can break down the forces acting on the suitcase into components.
1. Start by drawing a free-body diagram of the suitcase. The weight of the suitcase acts vertically downward with a magnitude of mg, where m is the mass of the suitcase (18 kg) and g is the acceleration due to gravity (approximately 9.8 m/s^2).
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2. Resolve the weight vector into two components. The vertical component is equal to mg, and the horizontal component is mg * sin(41°). The vertical component is balanced by the normal force, and the horizontal component is balanced by the tension in the strap.
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F_n | mg
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3. Since the suitcase is moving with constant speed, the net horizontal force must be zero. We can determine the frictional force acting on the suitcase using the coefficient of kinetic friction.
Frictional force (F_friction) = coefficient of kinetic friction * Normal force
= 0.31 * F_n
4. The tension in the strap is equal in magnitude but opposite in direction to the frictional force, as they balance each other horizontally. So, the tension in the strap is also 0.31 * F_n.
5. The normal force is equal in magnitude but opposite in direction to the vertical component of the weight. So, the normal force is equal to mg.
Therefore, the normal force is equal to mg (18 kg * 9.8 m/s^2) and the tension in the strap is 0.31 times the normal force.