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.
(a) Determine the horizontal component of the worker's push.
(b) Write a net force equation for the horizontal forces on the box.
1 answer:
Answer:
a) Fₓ = 23.5 N
b) Net force = Fₓ
Explanation:
An image of the question as described is attached to this solution.
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
Hope this Helps!!!
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Answer:
The magnitude of applied force,parallel to the incline is 575.38 N and parallel to the floor is 605 N.
Explanation:
Given:
Mass of the piano = 190 kg
Inclined angle = 18 degree
Considering gravity, = 9.8
And
Using, and
<em>FBD diagram is attached with all the force acting on the floor and and the inclined. </em>
We have to find the magnitude of forces,when the man pushes it parallel to the incline and to the floor.
a.
When the man pushes it parallel to the incline.
Balancing the forces as .
⇒
⇒
⇒ Here it is negative as the force is acting downward.
⇒ Plugging the values of mass and angle
.
⇒
⇒ N
b.
When the force is parallel to the floor.
⇒
⇒
⇒ Plugging the values.
⇒
⇒ N
So,
The magnitude of applied force in inclined direction is 575.38 Newton and parallel to the floor is 605 N.
