Forces on an object, Force vector

[Rileyss123]

There are two forces on the 1.1 kg box in the overhead view of Fig. 5-37 but only one is shown. The figure also shows the acceleration of the box.

IMAGE : http://www.webassign.net/hrw/05_37.gif

(a) Find the second force in unit-vector notation.
N i + N j

(b) Find the second force as a magnitude and direction.
N,
° (counterclockwise from the +x-axis is positive)

I'm not exactly sure where to start. Using F=ma then F= (1.1)(12)=13.2
so is the net Force 13.2
and for N in part be would the answer be either 20tan(30) or 20/tan(30) ??

 

[ritwik06]

There are two forces on the 1.1 kg box in the overhead view of Fig. 5-37 but only one is shown. The figure also shows the acceleration of the box.

IMAGE : http://www.webassign.net/hrw/05_37.gif

(a) Find the second force in unit-vector notation.
N i + N j

(b) Find the second force as a magnitude and direction.
N,
° (counterclockwise from the +x-axis is positive)

I'm not exactly sure where to start. Using F=ma then F= (1.1)(12)=13.2
so is the net Force 13.2
and for N in part be would the answer be either 20tan(30) or 20/tan(30) ??

You are asked to find the second force and not the net force.
As you have calculated the net force and you know its direction, resolve it into components
F(net)= -13.2 sin 30 i + -13.2 cos 30 j
F1= 20 i
F1+F2=F(net)

Find F2 now

 

[baba89]

I would approach this problem by first identifying the known variables and using the appropriate equations to solve for the unknown forces. In this case, we know the mass of the box (1.1 kg) and its acceleration (12 m/s^2). We also know that there are two forces acting on the box, one of which is shown in the figure.

To find the second force in unit-vector notation, we can use Newton's second law, F=ma, where F is the net force acting on the box. Since we know the mass and acceleration, we can plug these values into the equation and solve for F. However, we also need to consider the direction of the force. Since the acceleration is shown to be in the +x direction, we can assume that the net force is also in the +x direction. Therefore, the second force in unit-vector notation would be F = (1.1)(12) = 13.2 N i.

To find the second force as a magnitude and direction, we can use the Pythagorean theorem to find the magnitude of the force. The Pythagorean theorem states that the magnitude of the net force is equal to the square root of the sum of the squares of the individual forces. In this case, we know the magnitude of one force (13.2 N) and we need to find the magnitude of the second force. Using the Pythagorean theorem, we can solve for the magnitude of the second force, which is approximately 18.2 N.

To find the direction of the second force, we can use trigonometric functions. Since the second force is acting in the +x direction, we can use the tangent function to find the angle between the second force and the +x axis. The tangent of this angle can be found by dividing the y-component of the force (N j) by the x-component of the force (N i). Therefore, the tangent of the angle is 20/13.2 = 1.515. To find the angle itself, we can use the inverse tangent function, which gives us an angle of approximately 56.3°. This means that the second force is acting at an angle of 56.3° counterclockwise from the +x axis.

In conclusion, as a scientist, I would approach this problem by using known equations and principles to solve for the unknown forces and their magnitudes and directions. By identifying