Consider the skier on a slope shown in Figure 2. 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?

Figure 2: 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). NN size 12{N} {} is perpendicular to the slope and f is parallel to the slope, but ww size 12{w} {} has components along both axes, namely w⊥w⊥ size 12{w rSub { size 8{ ortho } } } {} and w∥ w∥ . NN size 12{N} {} is equal in magnitude to w⊥w⊥ size 12{w rSub { size 8{ ortho } } } {}, so that there is no motion perpendicular to the slope, but ff size 12{f} {} is less than w∥ 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 ww size 12{w} {}, ff size 12{f} {}, and NN size 12{N} {} in Figure 2. NN size 12{N} {} is always perpendicular to the slope, and ff size 12{f} {} is parallel to it. But ww 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∥ w∥ size 12{w rSub { size 8{ \lline \lline } } } {} to be the component of weight parallel to the slope and w⊥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∥ =wsin(25º)=mgsin(25º)w∥ =wsin(25º)=mgsin(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⊥=wcos(25º)=mgcos(25º)w⊥=wcos(25º)=mgcos(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∥ w∥ size 12{w rSub { size 8{ \lline \lline } } } {} and friction ff size 12{f} {}. Using Newton’s second law, with subscripts to denote quantities parallel to the slope,

where Fnet ∥ =w∥ =mgsin(25º)Fnet ∥ =w∥ =mgsin(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

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

and substituting this into Newton’s second law, a∥ =Fnet ∥ ma∥ =Fnet ∥ m size 12{a rSub { size 8{ \lline \lline } } = { {F rSub { size 8{"net " \lline \lline } } } over {m} } } {}, gives

We substitute known values to obtain

which yields

which is the acceleration parallel to the incline when there is 45.0 N of opposing friction.

Discussion

Since friction always opposes motion between surfaces, the acceleration is smaller when there is friction than when there is none. In fact, it is a general result that if friction on an incline is negligible, then the acceleration down the incline is a=gsinθa=gsinθ size 12{a=g"sin"θ} {}, regardless of mass. This is related to the previously discussed fact that all objects fall with the same acceleration in the absence of air resistance. Similarly, all objects, regardless of mass, slide down a frictionless incline with the same acceleration (if the angle is the same).

Calculate the tension in the wire supporting the 70.0-kg tightrope walker shown in Figure 6.

Figure 6: The weight of a tightrope walker causes a wire to sag by 5.0 degrees. The system of interest here is the point in the wire at which the tightrope walker is standing.

Strategy

As you can see in the figure, the wire is not perfectly horizontal (it cannot be!), but is bent under the person’s weight. Thus, the tension on either side of the person has an upward component that can support his weight. As usual, forces are vectors represented pictorially by arrows having the same directions as the forces and lengths proportional to their magnitudes. The system is the tightrope walker, and the only external forces acting on him are his weight ww size 12{w} {} and the two tensions TLTL size 12{T rSub { size 8{L} } } {} (left tension) and TRTR size 12{T rSub { size 8{R} } } {} (right tension), as illustrated. It is reasonable to neglect the weight of the wire itself. The net external force is zero since the system is stationary. A little trigonometry can now be used to find the tensions. One conclusion is possible at the outset—we can see from part (b) of the figure that the magnitudes of the tensions TLTL size 12{T rSub { size 8{L} } } {} and TRTR size 12{T rSub { size 8{R} } } {} must be equal. This is because there is no horizontal acceleration in the rope, and the only forces acting to the left and right are TLTL size 12{T rSub { size 8{L} } } {} and TRTR size 12{T rSub { size 8{R} } } {}. Thus, the magnitude of those forces must be equal so that they cancel each other out.

Whenever we have two-dimensional vector problems in which no two vectors are parallel, the easiest method of solution is to pick a convenient coordinate system and project the vectors onto its axes. In this case the best coordinate system has one axis horizontal and the other vertical. We call the horizontal the xx size 12{x} {}-axis and the vertical the yy size 12{y} {}-axis.

Solution

First, we need to resolve the tension vectors into their horizontal and vertical components. It helps to draw a new free-body diagram showing all of the horizontal and vertical components of each force acting on the system.

Figure 7: When the vectors are projected onto vertical and horizontal axes, their components along those axes must add to zero, since the tightrope walker is stationary. The small angle results in TT size 12{T} {} being much greater than ww size 12{w} {}.

Consider the horizontal components of the forces (denoted with a subscript xx size 12{x} {}):

The net external horizontal force Fnet x=0Fnet x=0 size 12{F rSub { size 8{"net x"} } = 0} {}, since the person is stationary. Thus,

Now, observe Figure 7. You can use trigonometry to determine the magnitude of TLTL size 12{T rSub { size 8{L} } } {} and TRTR size 12{T rSub { size 8{R} } } {}. Notice that:

Equating TL xTL x size 12{T rSub { size 8{"Lx"} } } {} and TRxTRx size 12{T rSub { size 8{"Rx"} } } {}:

Thus,

as predicted. Now, considering the vertical components (denoted by a subscript yy size 12{y} {}), we can solve for TT size 12{T} {}. Again, since the person is stationary, Newton’s second law implies that net Fy=0Fy=0 size 12{F rSub { size 8{y} } =0} {}. Thus, as illustrated in the free-body diagram in Figure 7,

Observing Figure 7, we can use trigonometry to determine the relationship between TLyTLy size 12{T rSub { size 8{L} rSub { size 8{y} } } } {}, TRyTRy size 12{T rSub { size 8{R} rSub { size 8{y} } } } {}, and TT size 12{T} {}. As we determined from the analysis in the horizontal direction, TL=TR=TTL=TR=T size 12{T rSub { size 8{L} } =T rSub { size 8{R} } =T} {}:

Now, we can substitute the values for TLyTLy size 12{T rSub { size 8{L} rSub { size 8{y} } } } {} and TRyTRy size 12{T rSub { size 8{R} rSub { size 8{y} } } } {}, into the net force equation in the vertical direction:

and

so that

and the tension is

Discussion

Note that the vertical tension in the wire acts as a normal force that supports the weight of the tightrope walker. The tension is almost six times the 686-N weight of the tightrope walker. Since the wire is nearly horizontal, the vertical component of its tension is only a small fraction of the tension in the wire. The large horizontal components are in opposite directions and cancel, and so most of the tension in the wire is not used to support the weight of the tightrope walker.