Question

Two ropes are attached a wagon one horizontal to the west with a tension force of 75N and other east and at an angle of 38 degrees upward from horizontal with a lension force of 125 N Find the components of the net force on the cart Show your work

Answer 1

Answer:

The components of the net force acting on the cart are;

Horizontal force, Fₓ ≈ 23.5 N acting east

The vertical force, $F_y$ ≈ 76.96 N

Explanation:

The parameters given in the question are;

The magnitude of the force on the rope acting horizontally to the west = 75 N

The magnitude of the force on the other rope east acting at an angle to the horizontal = 125 N

The angle to the horizontal of the force on the other rope east to the horizontal = 38°

Therefore, we have;

The horizontal component of the force acting east = 125 × cos(38°) ≈ 98.5 N

The net sum of the horizontal forces, ∑Fₓ noting that the horizontal forces are acting in opposite direction is given as follows;

∑Fₓ = 75 - 125 × cos(38°) = 75 - 98.3 = -23.5N

Given that the difference in forces is negative, and hat we took the force acting east as negative, we have that the net horizontal force is acting east with a horizontal component of 23.5N

The net vertical force, ∑F$_y$, is given as follows;

∑F$_y$ = 125 × sin(38°) ≈ 76.96 N

Therefore, the components of the net force acting on the cart are;

Horizontal force, Fₓ ≈ 23.5 N, the vertical force, $F_y$ ≈ 76.96 N

Answer 2

The components of the net force on the cart are 23.5 Newton and 76.96 Newton.

Given the following data:

• Tension force (horizontal) due west = 75 N

• Tension force (horizontal) due east = 125 N

• Angle of inclination = 38°

To find the components of the net force on the cart:

First of all, we would determine the horizontal component of the force acting due east.

$F_y = Fcos(\theta)\\\\F_y = 125cos(38)\\\\F_y = 125 \times 0.7880\\\\F_y = 98.5 \; Newton$

Next, we would find the resultant horizontal forces:

$\sum F_y = F + F_y\\\\\sum F_y = 75 + (- 98.5)\\\\\sum F_y = -23.5 \;Newton$

Therefore, the horizontal component of the net force acts in the opposite direction due east.

Horizontal component of the net force = 23.5 Newton

For the vertical component of the net force:

$\sum F_x = Fsin(\theta)\\\\\sum F_x = 125cos(38)\\\\ \sum F_x = 125 \times 0.6157\\\\ \sum F_x = 76.96\; Newton$

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