You\'re using a rope to pull a sled at constant velocity across level snow,with

Question

You're using a rope to pull a sled at constant velocity across level snow,with coefficient of kinetic friction 0.050 between sled and the snow. The sled is loaded with physics books, giving a total mass of 21kg. With what force should you pull with the rope horizontal?

Next, what would it be if the rope is at 30 degrees above horizontal?


Express your answers to two significant figures and include the appropriate units.

Explanation / Answer

a) Knowing that the sum of forces is equal to mass times the acceleration, ( F= m*a) we can find the answer. First, you said that the velocity was constant. This meens that the acceleration is 0. Sum of forces = m*0 = 0 Now imagine the sled [[[[[[[[[[[[[[[]]]]]]]]]]]]]]]] you have two forces ( -----> motion) the left force is the friction force, slowing down the sled. The right force is the one you apply to counter the decceleration of the friction. This means the two are equal ! Force(Friction) = Force(Applied) and by the coeefficient of kinetic friction we can find the Force(Friction) Coefficient*mass*gravity = (0.050)*(29kg)*(9,81) = Force(Friction) = Force applied !!! b) We are asked the same thing but this time the sled is in a 30degree angle. The same forces will be applied but there will be one more force: gravity Draw the sled on an inclined plane and draw the forces. When you pivot your axis, your FBD should look like this: ........../ ......../ ......V You see that the bottom force is in an angle of 30 degrees with the vertical line passing throught the sled. Now, still knowing that the sum of forces equals 0, find Force(Applied) by: Force applied = Force of friction + Force of gravity Because we only want the forces in the horizontal plane of the inclined plane we will only take the X portion of the force of gravity: So Force(Applied) = (coefficient of friction)*(Mass)*(Gravity)*cos(30) + (Mass)*(Gravity)*sin(30) And there you have it! Noticed that the force needed calculated in B is bigger than the force needed in A ... It makes sense, because in B we need to fight gravity !!! Also, notice the change in the Force of friction in the second part. This is because, as you might have learned, the friction force is the coefficient of friction times the normal force in A, the normal force is simply the weight in B, the normal force is different. As you see in your free body diagram, there is two forces in the y direction: the normal and the weight of the sled * cos(30) !

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