- #1
Why would you think that "Logically the req. force should be -3430N * cos(15)"? The ski tow has to overcome the force back down the slope of the hill and that is parallel to the "opposite side" to the 15 degree angle in the right triangle you have. "opposite side/hypotenus" is sine, not cosine!Helicobacter said:
HallsofIvy said:
The formula for calculating the force for a ski tow's performance on a slope is force = mass x acceleration. This means that the force required to pull a skier up a slope will depend on the mass of the skier and the acceleration of the ski tow.
The mass of the skier can be determined by weighing the skier before they get on the ski tow. It is important to include the weight of any equipment or clothing the skier is wearing in the calculation of their mass.
The acceleration of a ski tow on a slope can be affected by several factors, including the angle of the slope, the condition of the snow, and the power of the ski tow's motor. Wind resistance and friction can also play a role in the acceleration of the ski tow.
The steeper the slope, the more force will be required for the ski tow's performance. This is because the greater the angle of the slope, the more gravity will be pulling the skier down, requiring the ski tow to exert more force to counteract it and pull the skier up the slope.
Yes, there are a few ways to increase the performance of a ski tow on a slope without increasing the force. One way is to improve the efficiency of the ski tow, for example by reducing friction or increasing the power of the motor. Another way is to decrease the mass of the skier, which will require less force to pull them up the slope.
