1. Two forces are applied to a box on a frictionless surface. One force is 40 N [S]. The second force is 70 N [N30ºW]. What is the overall net force?
a) 40.6 N [S59.5ºE]
b) 110 N
c) 40.6 N [N59.5ºW]
d) 40.6 N [S30.5ºW]
2. When drawing free body diagrapms, it is important to:
a) draw your diagram to scale
b) add force vectors tip-to-tail
c) represent the magnitude force using the length of the arrows
d) all of the above
a. one force is south, the other is slightly north of west. Well, the resultant definitely is towards the west, which eliminates a, b.
So what is the N component of the second force Northcomponent=70sin30 = 35Newtons, but there is a 40N S force, so the net force is S of W
b. silly question,when drawing one never has it to scale, only approximately so. tip to tail is important.
what about magnitude? well it helps, but we always just sketch.
So b is the best answer, but I can see a poorly prepared teacher ready to argue. Don't argue with the teacher, it is a losing battle, usually, in Physics classes.
1. To solve this problem, we need to find the horizontal and vertical components of the second force:
Horizontal component: The given angle is N30ºW, which means the force is 30º west of north. We can find the horizontal component by multiplying the magnitude (70 N) by the cosine of the angle.
Horizontal component = 70 N * cos(30º) ≈ 60.50 N
Vertical component: Similarly, we can find the vertical component by multiplying the magnitude (70 N) by the sine of the angle.
Vertical component = 70 N * sin(30º) ≈ 35 N
Now, we can determine the net force by adding the horizontal components and vertical components separately. Since the surface is frictionless, there is no force opposing the motion in the horizontal direction. Therefore, the net force in the horizontal direction is simply the horizontal component of the second force:
Net horizontal force = 60.50 N
In the vertical direction, we have the force of 40 N acting north and the vertical component of the second force acting south (opposite direction of north). Therefore, the net vertical force is:
Net vertical force = 40 N - 35 N = 5 N [down]
Now, we can use the Pythagorean theorem to find the magnitude of the net force:
Net force = √((Net horizontal force)^2 + (Net vertical force)^2)
= √((60.50 N)^2 + (5 N)^2)
≈ 60.56 N
The direction of the net force can be found using trigonometry as well. The angle can be calculated as:
θ = arctan(Net vertical force / Net horizontal force)
= arctan(5 N / 60.50 N)
≈ 4.57º
Since the net vertical force is downward, the angle would be measured clockwise from the positive x-axis (east). Therefore, the overall net force is approximately:
Net force = 60.56 N [E4.57ºS]
Thus, the correct answer is not listed.
2. When drawing free body diagrams, it is important to:
d) all of the above
Drawing the free body diagram to scale helps in accurately representing the relative magnitudes of the forces involved. This can provide a clearer understanding of the forces acting on the object.
Adding force vectors tip-to-tail helps visualize how the forces interact and allows for the determination of the net force.
Representing the magnitude of forces using the length of the arrows in the free body diagram helps in understanding the relative strength of each force.
Therefore, all of the options given are important when drawing free body diagrams.