A sled is pulled across a horizontal snow surface with a constant velocity. The force being applied is 80 N at an angle of 53 degrees to the ground. What is the magnitude of the force of friction of the snow acting on the sled?
To find the magnitude of the force of friction acting on the sled, we need to understand the forces involved.
1. The applied force: The sled is being pulled with a force of 80 N at an angle of 53 degrees to the ground. This force can be broken down into its horizontal and vertical components.
- Horizontal component: This component of the force is responsible for overcoming the force of friction acting on the sled.
- Vertical component: This component of the force doesn't directly affect the horizontal motion of the sled and can be ignored for this calculation.
2. The force of friction: The force of friction is the force exerted by the snow in the opposite direction to the motion of the sled.
Since the sled is moving with a constant velocity, it means the applied force is equal to the force of friction.
To find the horizontal component of the applied force, we can use trigonometry.
Horizontal component of the force = Applied force * cos(angle)
Horizontal component = 80 N * cos(53 degrees)
Once we have the horizontal component of the applied force, we can determine the force of friction since it is the same magnitude but in the opposite direction.
So, the magnitude of the force of friction acting on the sled is 80 N * cos(53 degrees).
To find the magnitude of the force of friction of the snow acting on the sled, we can use the following steps:
Step 1: Resolve the force being applied into its horizontal and vertical components.
The horizontal component of the force can be found using the formula:
F_horizontal = Force * cos(angle)
F_horizontal = 80 N * cos(53°)
F_horizontal ≈ 80 N * 0.6018 ≈ 48.144 N
Step 2: Since the sled is moving at a constant velocity, the net force acting on it must be zero. This means that the force of friction is equal in magnitude and opposite in direction to the horizontal component of the applied force.
So, the magnitude of the force of friction is approximately 48.144 N.
Therefore, the magnitude of the force of friction of the snow acting on the sled is approximately 48.144 N.
ANGLE=ang
ang<80>/f(x>(l/2)(ff)(mg)
<260>(1.333%)(3.7)
%=fn/viv<0.16/.409n)
[pi^2](fn(80))
ANS: 41.19