One student is pushing on a chair with a force of 260.6 N directed at an angle of 30 degrees above horizontal while a second student pushes on the same side of the chair with a force of 21.0 N at an angle of 15 degrees below horizontal. What is the magnitude, in N, of the sum of these forces?

Question

One student is pushing on a chair with a force of 260.6 N directed at an angle of 30 degrees above horizontal while a second student pushes on the same side of the chair with a force of 21.0 N at an angle of 15 degrees below horizontal. What is the magnitude, in N, of the sum of these forces?

how do i go about solving this question?

physics - bobpursley, Thursday, October 1, 2009 at 5:39pm
add them.

using i,j coordinates

F= 260.6(sin30 j+ cos30 i)+ 21(sin-15 j+ cos 15 i)

Srry bob but i get a wacky number of number of 236434N when i did it just know.

Answer 1

Find your error.

Or, post your work, and I will check.

Answer 2

thanks for the quick reply.

Recorrection

F= 260.6(sin30 j+ cos30 i)+ 21(sin-15 j+ cos 15 i)

f= 5,4726( 0.24j + 1.8i)

f=5,4726(0.24)^2 + 1.8 ^2

f=18255.34N

Answer 3

To solve this question, you need to break down the forces into their horizontal and vertical components and then add them together.

First, let's break down the force exerted by the first student:
Force = 260.6 N
Angle = 30 degrees above horizontal

To find the horizontal component, you can use the equation:
Horizontal component = Force × cos(angle)

Horizontal component = 260.6 N × cos(30 degrees)
Horizontal component = 260.6 N × √3/2
Horizontal component = 225.628 N

To find the vertical component, you can use the equation:
Vertical component = Force × sin(angle)

Vertical component = 260.6 N × sin(30 degrees)
Vertical component = 260.6 N × 1/2
Vertical component = 130.3 N

Now let's break down the force exerted by the second student:
Force = 21.0 N
Angle = 15 degrees below horizontal

To find the horizontal component, you can use the equation:
Horizontal component = Force × cos(angle)

Horizontal component = 21.0 N × cos(-15 degrees)
Horizontal component = 21.0 N × cos(-15 degrees)
Horizontal component = 20.604 N

To find the vertical component, you can use the equation:
Vertical component = Force × sin(angle)

Vertical component = 21.0 N × sin(-15 degrees)
Vertical component = 21.0 N × -1/4
Vertical component = -5.25 N

Now that you have the horizontal and vertical components for both forces, you can add them together:

Horizontal total = Horizontal component of the first force + Horizontal component of the second force
Horizontal total = 225.628 N + 20.604 N
Horizontal total = 246.232 N

Vertical total = Vertical component of the first force + Vertical component of the second force
Vertical total = 130.3 N + (-5.25 N)
Vertical total = 125.05 N

To find the magnitude of the sum of these forces, you can use the Pythagorean theorem:
Magnitude = √(Horizontal total^2 + Vertical total^2)

Magnitude = √(246.232 N^2 + 125.05 N^2)
Magnitude = √(60731.952 N^2)
Magnitude ≈ 246.53 N

Therefore, the magnitude of the sum of these forces is approximately 246.53 N.

Answer 4

well, the first step to the second is wrong, there is no way 260+21N became 54726 Newtons.

I don't follow your last step (didn't punch it in the calc), but you have to take sqrt of the sum of forces, I don't see where you did that. Resultant= sqrt(xforeces + y forces)