Question

1. Two ropes are attached to a wagon, one horizontal to the west with a tension force of 20 N, and the other east and at an angle of 30° northward and a tension force of 50 N. Find the components of the net force on the cart. Show all your work.

Answer 1

The horizontal and vertical components of the net force of the given ropes are;

Horizontal component of Net force = 23.3 i^

Vertical component of net force = 25 j^

We are given;

Tension in Rope 1; T₁ = 20 N

Tension in Rope 2; T₂ = 50 N

      We are told that The tension in Rope 1 is horizontal to the east. Thus;

Horizontal component of T₁ = -20 i^

      The Tension in rope 2 is acting at an angle of 30° northward. Thus;

Horizontal component of T₂ = (50 cos 30)i^

Vertical component of T₂ = (50 sin 30)j^  

• Thus, Net force is;

Net force = Sum of vertical component + sum of horizontal components

F_net = (50 sin 30)j^  + ((50 cos 30)i^ + (-20 i^)

This will be simplified to get;

F_net = 25 j^ + (43.3 - 20)i^

F_net = 23.3 i^ + 25 j^

• In conclusion, the horizontal and vertical components of the net force are respectively 23.3 i^ and 25 j^

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Answer 2

The resultant force of the two ropes on the given cart is 68.1 N.

The given parameters;

• the first force = 20 N to west
• the second force = 50 N to east at 30°

The net horizontal component force on the cart is calculated as follows;

∑Fₓ = Fₓ₁  + Fₓ₂

∑Fₓ = 20 x cos(0)  +  50 x cos(30)

∑Fₓ = 20  +  43.3

∑Fₓ = 63.3 N

The net vertical component of the force on the cart;

$\Sigma F_y = 20\times sin (0) \ + \ 50 \times sin(30)\\\\\Sigma F_y = 25 \ N$

The resultant force on the cart is calculated as follows;

$F = \sqrt{F_x^2 +F_y^2} \\\\F = \sqrt{63.3^2 + 25^2} \\\\F = 68. 1 \ N$

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