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E) Nbt is equal to the y-component of the weight, so it is slightly less than 10 N, or (10 N) cos 5°. (f) Nbt is the same as in (d)—it doesn’t matter whether the applied force of 5 N downward is due to a book sitting on top of it or something else. From smallest to greatest, (b), (a), (e), (c), (d) = (f). 43 Chapter 2: Force College Physics 72. Strategy Since the crate is at rest, the net force on it must be zero. The force of static friction must oppose the force of gravity parallel to the ramp.
N Find an expression for the altitude, h. 371× 103 km) ⎜ − 1⎟ = 2639 km . ⎜ 12 ⎟ n ⎝ n ⎠ ⎝ ⎠ 39 Chapter 2: Force College Physics 61. Strategy Gravitational field strength is given by g = GM R 2 . Find H = R2 − R1 , where R1 and R2 are the distances from the center of the Earth to the surface of the Earth and the location of the balloon, respectively. Solution Find the height above sea level of the balloon, H. 803 N kg)−1/ 2 ⎤ = 4 km ⎣ ⎦ 62. Strategy The gravitational field strength is given by g = GM R 2 .
Because of this, the normal force exerted on the crate by the floor is equal in magnitude to the weight of the crate; and the force of static friction exerted on the crate by the floor is equal in magnitude to the force exerted on the crate by Phineas. The magnitudes of the forces exerted on the crate are: weight = normal force = 350 N; force exerted by Phineas = static friction = 150 N (e) Only forces acting on the crate are shown in the FBD, not interaction partners which are forces due to the crate; so no; these forces are equal and opposite because the net force on the crate is zero.