The mass is
m= (|Fthrust|sinθ+|Flift|cosθ)/g
= 7.0× 104 kg.Page 234, problem 64:
(a) Since the wagon has no acceleration, the total forces in both thexandydirections must
be zero. There are three forces acting on the wagon: T,Fg, and the normal force from the
ground,Fn. If we pick a coordinate system withxbeing horizontal andyvertical, then the
angles of these forces measured counterclockwise from thexaxis are 90◦−φ, 270◦, and 90◦+θ,
respectively. We have
Fx,total=Tcos(90◦−φ) +Fgcos(270◦) +Fncos(90◦+θ)
Fy,total=Tsin(90◦−φ) +Fgsin(270◦) +Fnsin(90◦+θ),which simplifies to
0 =Tsinφ−Fnsinθ
0 =Tcosφ−Fg+Fncosθ.The normal force is a quantity that we are not given and do not wish to find, so we should
choose it to eliminate. Solving the first equation forFn= (sinφ/sinθ)T, we eliminateFnfrom
the second equation,
0 =Tcosφ−Fg+Tsinφcosθ/sinθ
and solve forT, finding
T=
Fg
cosφ+ sinφcosθ/sinθ
Multiplying both the top and the bottom of the fraction by sinθ, and using the trig identity for
sin(θ+φ) gives the desired result,T=
sinθ
sin(θ+φ)Fgs(b) The case ofφ = 0, i.e. pulling straight up on the wagon, results inT =Fg: we simply
support the wagon and it glides up the slope like a chair-lift on a ski slope. In the case of
φ= 180◦−θ,T becomes infinite. Physically this is because we are pulling directly into the
ground, so no amount of force will suffice.
Page 234, problem 65:
(a) If there was no friction, the angle of repose would be zero, so the coefficient of static friction,
μs, will definitely matter. We also make up symbolsθ,mandgfor the angle of the slope, the
mass of the object, and the acceleration of gravity. The forces form a triangle just like the one
in example 68 on page 207, but instead of a force applied by an external object, we have static
friction, which is less thanμsFn. As in that example,Fs=mgsinθ, andFs< μsFn, so
mgsinθ < μsFn.From the same triangle, we haveFn=mgcosθ, somgsinθ < μsmgcosθ.