A skier of weight 700N is pulled at constant speed up a smooth slope, of inclination 15 degrees, by a force of magnitude P N acting at 25 degrees upwards from the slope. Find the value of P.
Since there is no acceleration or friction, the component of P along the direction of motion equals the component of the weight in the oppostie direction.
P cos25 = W sin15
Solve for P
W is the skier's weight in newtons
To find the value of P, we can analyze the forces acting on the skier.
First, let's resolve the force of P into two components: one parallel to the slope and one perpendicular to the slope.
The force P can be divided into two components, P_parallel and P_perpendicular.
P_parallel is the component of P parallel to the slope, and P_perpendicular is the component of P perpendicular to the slope.
Given that the slope has an inclination of 15 degrees, we can determine the values of P_parallel and P_perpendicular using trigonometric functions.
P_parallel = P * sin(25 degrees)
P_perpendicular = P * cos(25 degrees)
The skier is pulled up the slope at a constant speed, which means the net force acting on the skier is zero.
The net force acting in the direction parallel to the slope is the force of gravity component perpendicular to the slope, which can be calculated as:
Force_gravity = weight of the skier = 700 N
Force_gravity_parallel = Force_gravity * sin(15 degrees)
Since the skier is being pulled up the slope at a constant speed, the force of P_parallel must be equal in magnitude and opposite in direction to the force of gravity parallel to the slope.
Therefore, P_parallel = Force_gravity_parallel
Now, we can substitute the values we obtained earlier to solve for P:
P * sin(25 degrees) = Force_gravity * sin(15 degrees)
P = (Force_gravity * sin(15 degrees)) / sin(25 degrees)
P = (700 N * sin(15 degrees)) / sin(25 degrees)
Calculating this expression, we find P ≈ 507.4 N.
Therefore, the force P has a magnitude of approximately 507.4 N.