Suppose a friend asks to examine the 10.0 kg box , hoping to guess what is inside; and you respond, "Sure, pull the box by the attached cord, as shown in figure, along the smooth surface of the table.
The magnitude of the force exerted by the person is
F (p) = 40.0 N.
It is exerted at a 30 degree angle as shown in figure.
Calculate (a) the acceleration of the box, and (b) the magnitude of the upward force F(N) exerted by the table on the box. Assume that friction can be neglected.
Solution :
1. Draw a sketch : The situation is shown in Fig. (a) , it shows the box and the force applied by the person, F (p).
2. free-body diagram : Fig. (b) shows the free-body diagram of the box. To draw it correctly, we show all the forces acting on the box and only the forces acting on the box.
They are : the force of gravity mg ; the normal force exerted by the table F (N) ; and the force exerted by the person F(p).
3. Choose axes and resolve vectors : We expect the motion to be horizontal , so we choose the x axis horizontal and the y axis vertical . The pull of 40.0 N has components
F (px) = (40.0 N) ( cos 30 degree) = (40.0 N) ( 0.866) = 34.6 N
F (py) = (40.0 N) ( sin 30 degree) = (40.0 N) (0.500) = 20.0 N.
In the horizontal (x) direction, F (N) and mg have zero components.
The horizontal component of the net force is F(px) .
4. (a) Apply Newton's second law to determine the x component of the acceleration;
F (px) = ma (x).
5. (a) Solve :
a (x) = F(px) /m = (34.6 N)/ (10.0 kg) = 3.46 m/s^2.
The acceleration of the box is 3.46 m/s^2 to the right.
4. (b) Apply Newton's second law to the vertical (y) direction , with upward as positive:
F (y) = ma(y)
F (N) - mg + F (py) = ma (y).
5. (b) Solve: We have mg = (10.0 kg ) ( 9.80 m/s^2) = 98.0 N .
F (py) = 20.0 N.
F (py ) is less then mg, the box does not move vertically, so
a (y) = 0.
Thus, F (N) - 98.0 N + 20 N = 0.
F (N) = 78.0 N.
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