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Weight on an incline, a two-dimensional problem

Consider the skier on a slope shown in [link] . Her mass including equipment is 60.0 kg. (a) What is her acceleration if friction is negligible? (b) What is her acceleration if friction is known to be 45.0 N?

A skier is skiing down the slope and the slope makes a twenty-five degree angle with the horizontal. Her weight W, shown by a vector vertically downward, breaks into two components—one is W parallel, which is shown by a vector arrow parallel to the slope, and the other is W perpendicular, shown by a vector arrow perpendicular to the slope in the downward direction. Vector N is represented by an arrow pointing upward and perpendicular to the slope, having the same length as W perpendicular. Friction vector f is represented by an arrow along the slope in the uphill direction. IIn a free-body diagram, the vector arrow W for weight is acting downward, the vector arrow for f is shown along the direction of the slope, and the vector arrow for N is shown perpendicular to the slope.
Since motion and friction are parallel to the slope, it is most convenient to project all forces onto a coordinate system where one axis is parallel to the slope and the other is perpendicular (axes shown to left of skier). N size 12{N} {} is perpendicular to the slope and f is parallel to the slope, but w size 12{w} {} has components along both axes, namely w size 12{w rSub { size 8{ ortho } } } {} and w . N size 12{N} {} is equal in magnitude to w size 12{w rSub { size 8{ ortho } } } {} , so that there is no motion perpendicular to the slope, but f size 12{f} {} is less than w size 12{w rSub { size 8{ \lline \lline } } } {} , so that there is a downslope acceleration (along the parallel axis).

Strategy

This is a two-dimensional problem, since the forces on the skier (the system of interest) are not parallel. The approach we have used in two-dimensional kinematics also works very well here. Choose a convenient coordinate system and project the vectors onto its axes, creating two connected one -dimensional problems to solve. The most convenient coordinate system for motion on an incline is one that has one coordinate parallel to the slope and one perpendicular to the slope. (Remember that motions along mutually perpendicular axes are independent.) We use the symbols size 12{ ortho } {} and size 12{ \lline \lline } {} to represent perpendicular and parallel, respectively. This choice of axes simplifies this type of problem, because there is no motion perpendicular to the slope and because friction is always parallel to the surface between two objects. The only external forces acting on the system are the skier’s weight, friction, and the support of the slope, respectively labeled w size 12{w} {} , f size 12{f} {} , and N size 12{N} {} in [link] . N size 12{N} {} is always perpendicular to the slope, and f size 12{f} {} is parallel to it. But w size 12{w} {} is not in the direction of either axis, and so the first step we take is to project it into components along the chosen axes, defining w size 12{w rSub { size 8{ \lline \lline } } } {} to be the component of weight parallel to the slope and w size 12{w rSub { size 8{ ortho } } } {} the component of weight perpendicular to the slope. Once this is done, we can consider the two separate problems of forces parallel to the slope and forces perpendicular to the slope.

Solution

The magnitude of the component of the weight parallel to the slope is w = w sin ( 25º ) = mg sin ( 25º ) size 12{w rSub { size 8{ \lline \lline } } =w"sin" \( "25"° \) = ital "mg""sin" \( "25"° \) } {} , and the magnitude of the component of the weight perpendicular to the slope is w = w cos ( 25º ) = mg cos ( 25º ) size 12{w rSub { size 8{ ortho } } =w"cos" \( "25"° \) = ital "mg""cos" \( "25"° \) } {} .

(a) Neglecting friction. Since the acceleration is parallel to the slope, we need only consider forces parallel to the slope. (Forces perpendicular to the slope add to zero, since there is no acceleration in that direction.) The forces parallel to the slope are the amount of the skier’s weight parallel to the slope w size 12{w rSub { size 8{ \lline \lline } } } {} and friction f size 12{f} {} . Using Newton’s second law, with subscripts to denote quantities parallel to the slope,

a = F net m

where F net = w = mg sin ( 25º ) size 12{F rSub { size 8{"net " \lline \lline } } =w rSub { size 8{ \lline \lline } } = ital "mg""sin" \( "25"° \) } {} , assuming no friction for this part, so that

a = F net m = mg sin ( 25º ) m = g sin ( 25º )
( 9.80 m/s 2 ) ( 0.4226 ) = 4.14 m/s 2

is the acceleration.

(b) Including friction. We now have a given value for friction, and we know its direction is parallel to the slope and it opposes motion between surfaces in contact. So the net external force is now

Questions & Answers

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20/(×-6^2)
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it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
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ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
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Source:  OpenStax, Une: physics for the health professions. OpenStax CNX. Aug 20, 2014 Download for free at http://legacy.cnx.org/content/col11697/1.1
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