A 20 kg wagon is pulled along the level ground by a rope inclined at 30 deg above horizontal. A frictional force of 30 N opposes the motion, how large is the pulling force if the wagon is moving at constant velocity?
m*g = 20kg * 9.8N/kg = 196 N = Wt. of the wagon.
Fap*Cos30 - Fk = m*a
Fap*cos30-30 = m*0 = 0
Fap*cos30 = 30
Fap = 30/cos30o = 34.6 N. = Force applied.
Well, did you know that pulling a wagon takes a lot of strength, but sometimes you just need the right angle? In this case, since the wagon is moving at a constant velocity, it means that the net force acting on it is zero. So, the pulling force must balance out the frictional force.
We know that the frictional force is 30 N, so the pulling force must also be 30 N. Now, let's have a little math fun. Since the rope is inclined at 30 degrees above horizontal, we can use a little trigonometry to find the horizontal component of the pulling force.
The horizontal component of the pulling force is given by F = Fp * cos(theta), where Fp is the magnitude of the pulling force and theta is the angle between the rope and the horizontal.
In this case, theta is 30 degrees, so we can plug in these values and find:
F = 30 N * cos(30 deg) ≈ 26 N
So, the magnitude of the pulling force is approximately 26 N. And remember, when in doubt, just pull it out!
To determine the pulling force required for the wagon to move at a constant velocity, we need to consider the forces acting on the wagon.
Step 1: Identify the forces acting on the wagon:
- Gravity force (weight)
- Normal force
- Frictional force
- Pulling force
Step 2: Divide the forces into components:
The force of gravity can be divided into two components:
- Fg_parallel: This component acts parallel to the inclined plane (30-degree angle above the horizontal).
- Fg_perpendicular: This component acts perpendicular to the inclined plane.
Step 3: Calculate the force components:
The weight (Fg) of the wagon can be calculated using the formula: Fg = m * g
where m is the mass of the wagon (20 kg) and g is the acceleration due to gravity (9.8 m/s^2).
Fg = 20 kg * 9.8 m/s^2 ≈ 196 N
The force component parallel to the inclined plane (Fg_parallel) can be calculated using the formula: Fg_parallel = Fg * sin(30°).
Fg_parallel = 196 N * sin(30°) ≈ 98 N
The force component perpendicular to the inclined plane (Fg_perpendicular) can be calculated using the formula: Fg_perpendicular = Fg * cos(30°).
Fg_perpendicular = 196 N * cos(30°) ≈ 169.65 N
Step 4: Calculate the net force:
To determine the net force acting on the wagon, subtract the frictional force from the force parallel to the inclined plane.
Net force = Fg_parallel - frictional force
Net force = 98 N - 30 N = 68 N
Since the wagon is moving at a constant velocity, the net force must be zero. Therefore, the pulling force (F_pull) must be equal in magnitude but opposite in direction to the net force:
F_pull = - Net force
F_pull = - 68 N ≈ -68 N
Therefore, the pulling force required for the wagon to move at a constant velocity is approximately 68 N in the opposite direction of the net force.
To find the pulling force required to overcome the friction and maintain a constant velocity, we need to first analyze the forces acting on the wagon.
Let's break down the forces acting on the wagon:
1. Weight (mg): The weight of the wagon can be calculated by multiplying the mass (m) by the acceleration due to gravity (g). In this case, the weight is given by W = mg = (20 kg)(9.8 m/s²) = 196 N.
2. Normal force (N): The normal force is the force exerted by the ground on the wagon perpendicular to the inclined plane. Since the wagon is on level ground, the normal force is equal in magnitude and opposite in direction to the weight. Therefore, the normal force is N = 196 N.
3. Frictional force (F_friction): The frictional force opposes the motion of the wagon and has a magnitude of 30 N.
4. Pulling force (F_pull): This is the force exerted by the person pulling the wagon.
Now, let's analyze the forces in the horizontal direction:
The horizontal component of the weight does not affect the motion of the wagon because it is perpendicular to the direction of motion.
The horizontal component of the force of friction opposes the pulling force and can be calculated using the formula F_friction = μ × N, where μ is the coefficient of friction. However, the coefficient of friction is not provided in the question. To calculate the pulling force, we need to make an assumption about the coefficient of friction.
Let's assume that the coefficient of friction is μ = (F_friction / N) = (30 N / 196 N) = 0.153.
The sum of the horizontal forces is zero because the wagon is moving at a constant velocity. Therefore, the equation becomes:
F_pull - F_friction = 0.
Substituting the values for the frictional force and the coefficient of friction, we have:
F_pull - (0.153 × 196 N) = 0,
F_pull = 0.153 × 196 N,
F_pull ≈ 29.94 N.
Therefore, the magnitude of the pulling force required to maintain a constant velocity is approximately 29.94 N.