you slide a 325-N trunk up a 20.0 degree incline plane with a constant velocity by exerting a force of 211 N parallel to the incline plane
a)what is the component of the trunk's weight parallel to the plane ?
b)what is the sum of your applied force ,friction and the parallel component of the trunks weight?
c)what size and direction of the friction force?
d) what is the coefficient of friction?
a) 325sin20
b) sume is zero, it is at constant veloicty.
c) mg*mu*cos20, downhill (opposes motion)
d) 211-mg*mu(cos20)-325sin20=0
To answer these questions, we first need to understand the forces involved in this scenario.
1. Calculate the component of the trunk's weight parallel to the incline plane:
The weight of the trunk creates two components: one perpendicular to the incline (mg * cosθ) and one parallel to the incline (mg * sinθ). In this case, θ is the angle of the incline, which is 20.0 degrees. So, the parallel component of the trunk's weight is:
Weight_parallel = mg * sinθ
Weight_parallel = 325 N * sin(20.0°)
2. Calculate the sum of your applied force, friction, and the parallel component of the trunk's weight:
Since the trunk is moving with a constant velocity, the applied force must equal the force of friction. The sum of your applied force, friction, and the parallel component of the trunk's weight is:
Sum_of_forces = Applied_force + Force_of_friction + Weight_parallel
Sum_of_forces = 211 N + Force_of_friction + Weight_parallel
3. Determine the size and direction of the friction force:
Since the trunk is moving at a constant velocity, the force of friction must be equal in magnitude and opposite in direction to the applied force. So:
Force_of_friction = -Applied_force
Force_of_friction = -211 N
The negative sign indicates that the direction of the friction force is opposite to the applied force.
4. Calculate the coefficient of friction:
The coefficient of friction (μ) can be calculated using the equation:
μ = Force_of_friction / Normal_force
In this case, since the trunk is on an inclined plane, the normal force is not equal to the weight of the trunk (mg). To find the normal force, we need to calculate its component perpendicular to the incline:
Normal_force = mg * cosθ
Then, we can find the coefficient of friction:
Coefficient_of_friction = Force_of_friction / Normal_force
Coefficient_of_friction = (-211 N) / (325 N * cos(20.0°))
Note: In this case, the coefficient of friction will be negative, indicating that the applied force is greater than the force of friction, allowing the trunk to move at a constant velocity.
So, to summarize:
a) The component of the trunk's weight parallel to the plane is 112.2 N.
b) The sum of your applied force, friction, and the parallel component of the trunk's weight is 323.2 N.
c) The size of the friction force is 211 N, and its direction is opposite to the applied force.
d) The coefficient of friction is -0.652, indicating that the applied force is greater than the force of friction.
a) The component of the trunk's weight parallel to the plane can be found using the equation:
Weight_parallel = Weight * sin(angle)
Given:
Weight = 325 N,
Angle = 20.0 degrees,
Weight_parallel = Weight * sin(angle) = 325 N * sin(20.0) ≈ 110.83 N
b) The sum of the applied force, friction, and the parallel component of the trunk's weight is equal to zero, since the trunk is moving at a constant velocity. Therefore, the equation can be written as:
Applied force + Friction + Weight_parallel = 0
Given:
Applied force = 211 N,
Weight_parallel = 110.83 N,
Friction = - (Applied force + Weight_parallel) = - (211 N + 110.83 N) = -321.83 N
c) The friction force is equal to -321.83 N, which means it acts in the opposite direction of the applied force and the weight parallel to the plane.
d) The coefficient of friction (μ) can be calculated using the equation:
Coefficient of friction = Friction force / Normal force
However, we do not have the value of the normal force given in the question.