Forces -- Physics

Question

1. Four children are going to horizontally push a crate to set up a lemonade stand. One pushes with a force of 10.0 N [S], the second pushes with a force of 15.0 N [W], and the third pushes with a force of 12.0 N [W30°S]. The fourth defiant child who didn’t get his/her way decided to push the crate the other way.

a) Draw a free-body diagram including components

b) What force did the fourth child push with to prevent the crate from moving?

Answer 1

The Free-Body diagram would look like this:

Where the F₁ is the 10 N force, F₂ is the 15 N force, F₃ is the 12 N force and F₄ is the opposing force from the fourth child. Both F3 and F4 have components in the x and y directions as shown in dotted lines with the associated subscripts.

F_3x = F₃cos(30°) = 12cos(30°) = -10.3923 N

F_3y = F₃sin(30°) = 12sin(30°) = - 6.00 N

The sum of the forces in the y direction are then F₁ + F_3y = -10 + -6 = -16.0 N

The sum of the forces in the x direction are then F₂ + F_3x = -15 + -10.3923 = -25.3923 N

The magnitude of the Resultant Force is then √(16² + 25.3923²) = 30.0 N

The direction of the Resultant Force is tan^-1(16/25.3923) = W 32.2° S

To counteract this resultant force, child 4 must push with an equal magnitude (30.0 N) at an angle 180° off so at E 32.2° N (shown as angle θ in the free-body diagram)