Make a sketch Problem: A 10.0 kg box is pulled along a horizontal surface by a rope that makes a 30.0 o angle with the horizontal. The tension in the rope.

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Presentation transcript:

Make a sketch Problem: A 10.0 kg box is pulled along a horizontal surface by a rope that makes a 30.0 o angle with the horizontal. The tension in the rope is 30.0 N. There is 10.0 N of friction acting on the box. Find: a) the acceleration of the box b) The normal force of the ground 30 o

Make a sketch Draw a free-body diagram for each object mg n FAFA Problem: A 10.0 kg box is pulled along a horizontal surface by a rope that makes a 30.0 o angle with the horizontal. The tension in the rope is 30.0 N. There is 10.0 N of friction acting on the box. Find: a) the acceleration of the box b) The normal force of the ground f

Make a sketch Draw a free-body diagram for each object Choose axes parallel and perpendicular to the direction of motion mg n FAFA x y Problem: A 10.0 kg box is pulled along a horizontal surface by a rope that makes a 30.0 o angle with the horizontal. The tension in the rope is 30.0 N. There is 10.0 N of friction acting on the box. Find: a) the acceleration of the box b) The normal force of the ground f

Make a sketch Draw a free-body diagram for each object Choose axes parallel and perpendicular to the direction of motion Draw components for any vector not along one of the axes mg n FAFA x y F Ax = F A cos30 o F Ay = F A sin30 o F Ay F Ax Problem: A 10.0 kg box is pulled along a horizontal surface by a rope that makes a 30.0 o angle with the horizontal. The tension in the rope is 30.0 N. There is 10.0 N of friction acting on the box. Find: a) the acceleration of the box b) The normal force of the ground f Apply  F = ma in each dimension.

How to Apply  F = ma in each dimension: In any dimension where there is no acceleration: Set the sum of all the forces in one direction In any dimension where there is acceleration: Add up all the forces in the direction of the acceleration mg n FAFA x y F Ay F Ax f equal to the sum of all the forces in the opposite direction and subtract all the forces opposite to the acceleration. Set the difference equal to the mass times the acceleration.

F Ay F Ax mg n FAFA x y F A cos30 o - f = ma x F A sin30 o + N - mg = ma y (but a y = 0!) F A sin30 o + N = mg F Ax = F A cos30 o F Ay = F A sin30 o a x = (F A cos30 o – f )/m a x = [(30 N)(0.866) – 10.0 N] /(10 kg) a x = 1.60 m/s 2 N = mg - F A sin30 o N = (10 kg)(9.8 m/s2) - (30 N)(0.5) N = 83.0 N  F y = ma y  F x = ma x f