Two forces pull on ropes attached to a cart: 1 with a force of 200 N South, and the other with a force of 150 N at 25º W of S. What was the net force on the cart?

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justaguide's profile pic

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

The net force on the cart has to be calculated by adding the two forces that are acting on the cart. As force is a vector quantity, we need to perform vector addition.

The first force of 200 N pulls the cart towards the south.

The second force of 150 N, pulls the cart in a direction 25 degrees west of south. The component of the force towards the south is 150*cos 25 = 135.94 N. The component towards the west is 150*sin 25 = 63.39 N.

The total force towards the south is 335.94 N and the force towards the west is 63.39 N.

This gives the magnitude of the net force as sqrt(335.94^2 + 63.39^2) = 341.86 N acting in the direction arc tan(63.39/335.94) = 10.68 degrees west of south.

The net force of 341.86 N acts in a direction 10.68 degrees west of south.

pagu's profile pic

pagu | (Level 2) eNoter

Posted on

To sole the problem we

let F1 = 200N, South

F2 = 150N, 25º West of South

R the Net Force

R² = Fx² + Fy²

where the x-axis along the East-West and the y-axis along the Meridian line

Assuming that component force towards the N & E as positive and component force towards S & W as negative, then

For F1:

Fx1 = 0

and

Fy1 = -200N

For F2:

Fx2 = -150sin25º

Fx2 = -63.39N

and

Fy2 = -150cos25º

Fy2 = -135.95N

Fx = Fx1 + Fx2

Fx = -63.39 N

Fy = Fy1 + Fy2

Fy = -200 + (-135.95)

Fy = -335.95 N

R² = (-63.38)² + (-335.95)²

R = 341.87N

The direction (θ) of the net force can be computed by

tan θ = |Fx/Fy|

tan θ = |-63.39/-335.95|

Note (-) Fx means the direction is West and (-) Fy means the direction is South, then

θ = Arctan|-63.38/-335.95|

θ = 10º30'17.5" South-West

Therefore the net force is 341.87N at 10º30'17.5" South-West

 

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