Question

A bag is pushed with a force of 80.00 N
directed along the handle, which makes an
angle of 36° with the ground. Find the
horizontal and vertical components of the force
on the bag.

Answer 1

Answer:

Horizontal component is 64.7 N

vertical component is 47 N

Explanation:

As we know that if a vector makes some angle with the horizontal then its two components are given as

$A_x = A cos\theta$

$A_y = A sin\theta$

now we have

$A = 80 N$

$\theta = 36 degree$

now we have

$F_x = 80 cos36$

$F_x = 64.7 N$

Similarly we have

$F_y = 80 sin36$

$F_y = 47 N$

Answer 2

The horizontal and the vertical components of the force on the bag are 64.72 N and 47.04 N respectively.

To calculate the horizontal and the vertical component of the force, we use the equation below.

Formula:

• Fy = Fsin∅................... Equation 1
• Fx = Fcos∅..................... Equation 2

Where:

• Fy = vertical component of the force
• Fx = horizontal component of the force
• F = Force applied

• ∅ = Angle the force make to the ground.

From the question,

Given:

• F = 80 N
• ∅ = 36°

Substitute these values into equation 1 and equation 2 respectively.

• Fy = 80(sin36)
• Fy = 80(0.588)
• Fy = 47.04 N

Also for the horizontal component,

• Fx = 80(cos36)
• Fx = 80(0.809)
• Fx = 64.72 N.

Hence, The horizontal and the vertical components of the force on the bag are 64.72 N and 47.04 N respectively.

Learn more about components of forces here: https://brainly.com/question/12980489