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vizakenjack
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A loaded sled weighing 80N rests on a plane inclined at angle θ= 20° to the horizontal (see pic). Between the sled and the plane, the coefficient of static friction is 0.25, and the coefficient of kinetic friction is 0.15
Taken from http://www.uh.edu/%7Ewkchu/courses/phys1321/chapter6/Home%20Work%206%20Solutions.pdf , problem #16.
a) What is the least magnitude of the force parallel to the plane, that will prevent the sled from slipping down the plane?
b) What is the minimum magnitude F that will start the sled moving up the plane?
Now, I see what the solutions are, but I'm just trying to make sense of them.
So I decided to break up the object's force acting downwards (mg) into x and y components. x = sin(20°) * mg, y =
cos(20°) * mg. In other words, this is how the object's downward force acts in each of those x and y directions.Since the object is on an incline, normal force isn't simply equal to the magnitude of the object's downward force due to gravity in opposite direction (-mg), in this case, normal force - Fn, would only experience an object's partial downward force, which is as you can see from the graph is cos(20°) * mg, which is equivalent to the object's y component of the downward force due to gravity.
So the first question basically asks you how much force is necessary to apply to the object up the incline for it to remain at rest.
Let's see, the normal force (Fn) already takes care of the y component of the object's downward force generated by gravity. So you'd only need to overcome the remaining x component of the object's downward force (sin(20°) * mg), correct? At this point, friction is out of the calculations because the object doesn't need to be moving, but rather remain at rest / stationary on the incline.
But for some reason, they add up the friction force to the Force that needs to be applied, why? It's as if "friction" is helping us when we apply force to the object... which doesn't make sense. Friction does the opposite, when we apply force, our force needs to OVERCOME friction in order to get the object moving up an incline.
Can someone elaborate on this?
a) What is the least magnitude of the force parallel to the plane, that will prevent the sled from slipping down the plane?
b) What is the minimum magnitude F that will start the sled moving up the plane?
Now, I see what the solutions are, but I'm just trying to make sense of them.
So I decided to break up the object's force acting downwards (mg) into x and y components. x = sin(20°) * mg, y =
cos(20°) * mg. In other words, this is how the object's downward force acts in each of those x and y directions.Since the object is on an incline, normal force isn't simply equal to the magnitude of the object's downward force due to gravity in opposite direction (-mg), in this case, normal force - Fn, would only experience an object's partial downward force, which is as you can see from the graph is cos(20°) * mg, which is equivalent to the object's y component of the downward force due to gravity.
So the first question basically asks you how much force is necessary to apply to the object up the incline for it to remain at rest.
Let's see, the normal force (Fn) already takes care of the y component of the object's downward force generated by gravity. So you'd only need to overcome the remaining x component of the object's downward force (sin(20°) * mg), correct? At this point, friction is out of the calculations because the object doesn't need to be moving, but rather remain at rest / stationary on the incline.
But for some reason, they add up the friction force to the Force that needs to be applied, why? It's as if "friction" is helping us when we apply force to the object... which doesn't make sense. Friction does the opposite, when we apply force, our force needs to OVERCOME friction in order to get the object moving up an incline.
Can someone elaborate on this?
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