A worker pushes a 7 kg shipping box along a roller track. Assume friction is small enough to be ignored because of the rollers. The worker's push is 25 N directed down and to the right at an angle of 20°.
(a) Determine the horizontal component of the worker's push.
(b) Write a net force equation for the horizontal forces on the box.

From the image attached, the forces acting on the box include the weight of the box, the normal reaction of the surface on the box, the applied force on the box and the Frictional force opposing the motion of the box (which is negligible and equal to 0)

a) From the diagram, the horizontal component of the force is

Fₓ = 25 cos 20° = 23.49 N = 25 N

b) Again, from the diagram attached, doing a force balance on the box, in the horizontal direction, we obtain

Net force = Fₓ - Frictional force

But frictional force is 0 N

Net force = Fₓ

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onyebuchinnaji onyebuchinnaji · Answer 2 · 2021-12-03T16:58:46+01:00

(a) The horizontal component of the worker's push is 23.5 N.

(b) The net force equation for the horizontal force on the box is [tex]25 \ cos(20) = 7a[/tex]

The given parameters:

mass of the box, m = 7 kg
applied force on the box, F = 25 N
inclination of the force, θ = 20⁰

The horizontal component of the worker's push is calculated as follows;

[tex]F_x = F \times cos (\theta)\\\\F_x = 25 \times cos(20)\\\\F_x = 23.5 \ N[/tex]

The net force equation for the horizontal force on the box is calculated as follows;

[tex]\Sigma F = ma\\\\F\ cos(\theta) = ma\\\\25 \ cos(20) = 7 a[/tex]

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