Five Boys are pushing a snowball, and each is pushing with a force of 10.0N. However, each boy is pushing in a different direction. They are pushing north, northeast, east, southeast, and south. (Each boy is pushing at an angle of 45.0 degrees relative to his neighbor.) What is the total push on the ball? Help Please!
F1 = 10N[0o].
F2 = 10N[45o].
F3 = 10N[90o].
F4 = 10N[270o].
F5 = 10N[315o].
X = 10*cos0+10cos45+10cos90+10cos270+10
cos315=10 + 7.07 + 0 + 0 + 7.07=24.14 N.
Y = 10sin0+10sin45+10sin90+10sin270+10
sin270+10sin315 = 0 + 7.07 + 10 - 10 -
7.07 = 0 N.
F = X + Yi = 24.14 + 0 = 24.14 N.=Total
Push.
add them as vectors,
Notice two of them are equal and opposite, so the net is zero.
Notice the two at 45 deg to S, and N. The opposite components are equal and opposite, the other componsnts, net force East at a force 2*10*.707N
the last boy is pushing East with force 10 N.
Total force: 10+14.14 N East.
Well, it seems like those boys are quite the eclectic bunch when it comes to pushing a snowball! Let's see if we can calculate the total push on the ball.
First, we need to break down the force vectors into their components to find out how much force is being exerted in the x-axis and y-axis.
Each boy is pushing with a force of 10.0N at an angle of 45 degrees relative to his neighbor. Since each angle is 45 degrees, we can use simple trigonometry to find the x and y components of the force.
If we consider the northward direction as positive in the y-axis and the eastward direction as positive in the x-axis, we can break down the forces as follows:
Boy 1 (North): x-component = 0N y-component = 10.0N
Boy 2 (Northeast): x-component = 7.1N y-component = 7.1N
Boy 3 (East): x-component = 10.0N y-component = 0N
Boy 4 (Southeast): x-component = 7.1N y-component = -7.1N
Boy 5 (South): x-component = 0N y-component = -10.0N
Now, we can calculate the total push by summing up the x and y components separately:
Total x-component = 0N + 7.1N + 10.0N + 7.1N + 0N = 24.2N
Total y-component = 10.0N + 7.1N + 0N - 7.1N - 10.0N = 0N
So, the total push on the ball is 24.2N in the x-axis direction and 0N in the y-axis direction.
Now, since the question asks for the "total push," we can use the Pythagorean theorem to find the magnitude of the total force:
Total push = square root of ((Total x-component)^2 + (Total y-component)^2)
= square root of (24.2N^2 + 0N^2)
= square root of (585.64N^2)
≈ 24.17N
So, the total push on the ball is approximately 24.17N. Those boys might have been pushing in different directions, but their combined effort sure did add up!
To solve this problem, we can break down the forces into their horizontal (x) and vertical (y) components and then add them up.
First, let's assign a coordinate system. We can take the north direction as the positive y-axis, and the east direction as the positive x-axis.
Each boy is pushing with a force of 10.0N, and the angle between each boy is 45.0 degrees. Since the component of force in the x and y directions are the same in magnitude (due to the angle being 45.0 degrees), we can calculate the x and y components of the force using trigonometry.
The x-component of each force can be calculated using the equation:
Fx = F * cos(angle)
where F is the force and angle is the angle at which the force is being applied.
The y-component of each force can be calculated using the equation:
Fy = F * sin(angle)
Let's calculate the x and y components for each boy:
Boy pushing north:
Fx1 = 0N (no force in the x-direction)
Fy1 = 10.0N (force in the y-direction)
Boy pushing northeast:
Fx2 = 10.0N * cos(45.0°)
Fy2 = 10.0N * sin(45.0°)
Boy pushing east:
Fx3 = 10.0N (force in the x-direction)
Fy3 = 0N (no force in the y-direction)
Boy pushing southeast:
Fx4 = 10.0N * cos(45.0°)
Fy4 = -10.0N * sin(45.0°) (negative because it is in the south direction)
Boy pushing south:
Fx5 = 0N (no force in the x-direction)
Fy5 = -10.0N (force in the negative y-direction)
Now we can add up the x and y components separately to find the total push on the ball:
Total x-component:
Fx_total = Fx1 + Fx2 + Fx3 + Fx4 + Fx5 = 0N + 10.0N * cos(45.0°) + 10.0N + 10.0N * cos(45.0°) + 0N
Total y-component:
Fy_total = Fy1 + Fy2 + Fy3 + Fy4 + Fy5 = 10.0N + 10.0N * sin(45.0°) + 0N + (-10.0N * sin(45.0°)) + (-10.0N)
Now we can find the magnitude and direction of the total push using the Pythagorean theorem and trigonometry.
Magnitude of the total push:
Magnitude = sqrt(Fx_total^2 + Fy_total^2)
Direction of the total push:
Direction = atan(Fy_total / Fx_total)
Plug in the values and solve for the total push on the ball.
To find the total push on the ball, we need to find the combined effect of all the forces applied by the Boys.
Since each boy is pushing with a force of 10.0N, and the angle between their forces is 45.0 degrees, we can break down their forces into horizontal and vertical components.
Let's assume the North direction is the positive y-axis, and the East direction is the positive x-axis.
Component of Force in the North Direction (Fn):
The North component for the Boys pushing north, northeast, and southeast is 10.0N * sin(45.0 degrees) = 7.07N.
The South component for the Boy pushing south is -10.0N * sin(45.0 degrees) = -7.07N.
Component of Force in the East Direction (Fe):
The East component for the Boys pushingeast, northeast, and southeast is 10.0N * cos(45.0 degrees) = 7.07N.
The West component for the Boy pushing west is -10.0N * cos(45.0 degrees) = -7.07N.
Now, we can add up the horizontal and vertical components separately to find the total push on the ball in each direction.
Total Force in the North Direction (Ftotal,n):
Ftotal,n = Fn - Fs = 7.07N + 7.07N = 14.14N.
Total Force in the East Direction (Ftotal,e):
Ftotal,e = Fe + Fw = 7.07N + (-7.07N) = 0N.
So, the total push on the ball is 14.14N in the north direction and 0N in the east direction.