A football coach sits on a sled while two of his players build their strength by dragging the sled across the field with ropes. The friction force on the sled is 800 N and the angle between the two ropes is 20°. How hard must each player pull to drag the coach at a steady 2.0 m/s?
I see the error. Then angle between the ropes is 20, but the angle between the rope and the resultant force is 10.
Tension*cos10=400
Since they want a steady velocity, then the sum of all forces is 0. Which means that the resultant of the two ropes is in equilibrium with the friction force.
To find out how hard each player must pull to drag the coach at a steady 2.0 m/s, we can use the concept of vector components.
First, let's calculate the horizontal and vertical components of the force. Since the coach is moving at a steady 2.0 m/s, the horizontal component of the force must cancel out the friction force acting on the sled. Therefore, the horizontal component of the force is equal to the friction force, which is 800 N.
Now, let's calculate the vertical component of the force. The vertical component is determined by the angle between the two ropes. Since the angle is 20°, we can use trigonometry to find the vertical component of the force.
The vertical component of the force can be calculated using the formula:
Vertical component = Force x sin(angle)
Vertical component = 800 N x sin(20°)
Vertical component ≈ 271.2 N
Therefore, each player must pull with a force of approximately 271.2 N in order to drag the coach at a steady 2.0 m/s.
To determine how hard each player must pull to drag the coach at a steady speed of 2.0 m/s, we can start by analyzing the forces acting on the sled.
First, we need to resolve the force into components. We can split the force into two perpendicular components, one parallel to the direction of motion (horizontal) and one perpendicular to the direction of motion (vertical). The vertical component does not affect the motion in this case since the coach is not accelerating vertically. Therefore, we will focus on the horizontal component, which is responsible for moving the coach.
The horizontal force acting on the sled is equal to the force of friction. We can find this force using the equation:
Friction force = coefficient of friction * normal force
In this case, the friction force is given as 800 N. However, we need to find the normal force, which is the force exerted by the field on the sled that acts perpendicular to the field surface.
Since the coach sits on the sled, his weight creates a normal force opposing the weight of the sled. Therefore, we can say:
Normal force = weight of coach
Now, we need to calculate the vertical force acting on the sled, which is the weight of the coach. The weight of the coach can be found using the equation:
Weight = mass * acceleration due to gravity
Assuming an average mass of a football coach to be around 100 kg and using the acceleration due to gravity as 9.8 m/s^2, we can calculate the weight of the coach:
Weight = 100 kg * 9.8 m/s^2 = 980 N
Now that we have the weight of the coach and the friction force, we can calculate the normal force:
Normal force = 980 N
Next, we can calculate the horizontal component of the force exerted by each player. Since the two ropes form an angle of 20°, each player will exert a force at a 20° angle to the horizontal.
By using the trigonometric relationship of cosine, we can determine the horizontal component of the force exerted by each player:
Force exerted by each player = horizontal component of the force = friction force = 800 N
Finally, to calculate the force exerted by each player, we divide the horizontal component by the cosine of the angle:
Force exerted by each player = 800 N / cos(20°) = 842 N
Therefore, each player must pull with a force of approximately 842 N to drag the coach at a steady speed of 2.0 m/s.