The net force calculation involves understanding how opposing forces interact. In this scenario, a 10 N force north and a 50 N force south result in a net force of 40 N south, as the northward force is weaker. Additionally, a 40 N force west combines with the 40 N south force, forming a right triangle. Using the Pythagorean theorem, the resultant force is approximately 56.5 N directed southwest. Therefore, the initial claim of 100 N as the net force is incorrect.
#1
espo
the below forces act on an object; 10N north; 50 N south, and 40N west. what is themagnitude of the net force? my answer is 100 n is that correct?
Think about what it means for forces to oppose each other. Let's say that you tie two pieces of rope to a bowling ball, and you and your friend tug at the ends of the rope, on opposite sides of the bowling ball. As long as you pull equally hard as your friend, the bowling ball doesn't move. If you're pulling with 40 N of force in one direction, while your friend is pulling with 40 N of force in the other direction, the net force is zero. The forces, directed in opposite directions, cancel each other out, and the ball does not move.
In the case you've given, the north and south pulls are opposite, and so cancel. They don't cancel fully, however, because they are of different strengths; the person on the south side is pulling with 40 N more force than is the person on the north.
In other words, when you have a 10 N force to the north and a 50 N force to the south, it's just the same as having a 40 N force to the south and no force at all coming from the north.
So you can combine the 10 N north and 50 N south forces into one -- by adding them -- and get one 40 N south force.
Now what is the combination -- the sum -- of a 40 N south force and a 40 N west force?
Draw a little picture of the bowling ball, with the forces being applied to it in different directions represented by arrows, like this:
Code:
40 N
<---------*
|
|
| 40 N
|
v
The resulting force, the sum of these two forces, can be found by placing the arrows head-to-tail, like this:
Code:
40 N
<---------*
. |
. |
. | 40 N
. |
v v
Here, I've slid the south arrow to the left, so it's tail is on the head of the west arrow. This doesn't change the meaning of the arrow -- you can slide them around anywhere you'd like. Since they will always point in the same direction and have the same length, their meanings are not changed by sliding them around.
Now, we've got two arrows stacked head-to-tail. The first one leaves the origin, and goes a distance to the west. The second leaves the head of the first and goes some distance south. The sum of these two arrows is the arrow that connects the origin to the end of the last arrow. In other words, it's the slanted arrow shown here:
Code:
40 N
<·········*
. /
40 N . /
. /
. /
v/
Now, you should recognize this as a simple right triangle, with two short sides of length 40. What is the length of the hypotenuse? You can use the Pythagorean relationship:
c2 = a2 + b2
Plugging in 40 for A and 40 for B, you should find that the length of the resulting arrow has a length of about 56.5. Since in this drawing, length represents the magnitude of a force, you can conclude that the resulting force is about 56.5 Newtons. Furthermore, you can conclude that the resulting force is in the south-west direction. In compass degrees, that direction is 225 degrees.
Does this make sense?
- Warren
#3
M98Ranger
1,668
0
Yes, your answer is correct. The magnitude of the net force is the total sum of all the forces acting on the object. In this case, it is 10N + 50N + 40N = 100N.
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook.
Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water.
I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system
$$M(t) = M_{C} + m(t)$$
$$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$
$$P_i = Mv + u \, dm$$
$$P_f = (M + dm)(v + dv)$$
$$\Delta P = M \, dv + (v - u) \, dm$$
$$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$
$$F = u \frac{dm}{dt} = \rho A u^2$$
from conservation of momentum , the cannon recoils with the same force which it applies.
$$\quad \frac{dm}{dt}...
