What Forces Are Needed to Move a Sled Up an Inclined Plane?

[DJWise]

Friction up a slope (HELP!)

1. A sled weighing 80N rests ona plane inclines at angle 20 degrees to the horizontal. Between the sled and the plane, the coefficient of static friction is .25 and of kinetic friction is .15. A.) what is the least magnitude of the force F parrallel to the flane that will preven the sled from slipping down the plane? B.) What is the minimum mag. of F that will start the sled moving up the plane? C.) What value of F is required to move the sled up the plane at constant velocity?



2. For a I made friction and weight directed down the slope and the Force directed up the slope. I said that the Force had to be greater than the sum of friction and weight directed down the slope. Am I right in my thinking. I don't know how to tackle b and c. For c i got the same answer because accl would equal zero but it doesn't seem logical.

 

[LowlyPion]

1. A sled weighing 80N rests ona plane inclines at angle 20 degrees to the horizontal. Between the sled and the plane, the coefficient of static friction is .25 and of kinetic friction is .15. A.) what is the least magnitude of the force F parrallel to the flane that will preven the sled from slipping down the plane? B.) What is the minimum mag. of F that will start the sled moving up the plane? C.) What value of F is required to move the sled up the plane at constant velocity?

2. For a I made friction and weight directed down the slope and the Force directed up the slope. I said that the Force had to be greater than the sum of friction and weight directed down the slope. Am I right in my thinking. I don't know how to tackle b and c. For c i got the same answer because accl would equal zero but it doesn't seem logical.

In a) the force you need to supply is in addition to the resistance of friction. (Friction helps hold it.) So actually then the Force 80 N down is paid for with the friction AND the force you supply. It's in b) that you have to over come the friction as well as the weight component down the slope.

Once moving in c) you only have to equal the weight and the kinetic friction resistance to maintain unaccelerated velocity.

 

[baba89]

I can provide some insights and calculations to help you understand the concept of friction up a slope.

Firstly, it is important to understand that friction is a force that opposes motion between two surfaces in contact. In the case of a sled on a slope, there are two types of friction at play - static friction and kinetic friction.

Static friction is the force that keeps the sled from slipping down the slope when it is at rest. This force is proportional to the normal force (perpendicular to the surface) and the coefficient of static friction. In this case, the normal force is equal to the weight of the sled, which is 80N. Therefore, the maximum static friction force is (0.25)(80N) = 20N.

Kinetic friction, on the other hand, is the force that opposes the motion of the sled as it slides down the slope. This force is proportional to the normal force and the coefficient of kinetic friction. In this case, the kinetic friction force is (0.15)(80N) = 12N.

Now, let's answer the questions one by one:

A) To prevent the sled from slipping down the slope, the force parallel to the plane (F) must be equal to or greater than the maximum static friction force. Therefore, the minimum magnitude of F is 20N.

B) To start the sled moving up the slope, the force parallel to the plane must overcome the force of kinetic friction. This means that the force must be greater than the sum of kinetic friction and weight directed down the slope. Therefore, the minimum magnitude of F is (12N + 80N) = 92N.

C) To move the sled up the plane at a constant velocity, the force parallel to the plane must be equal to the force of kinetic friction. This means that the force must be equal to the sum of kinetic friction and weight directed down the slope. Therefore, the magnitude of F required is (12N + 80N) = 92N.

In conclusion, your thinking for part A and B is correct. However, for part C, the force required to move the sled up the slope at a constant velocity is equal to the force of kinetic friction, not the sum of kinetic friction and weight. I hope this explanation helps you understand the concept of friction up a slope better. Keep up the good work in your studies!