PHYSICS: Please help :) Thank you! 1. A crane pulls up on a bin with a mass of 250kg with a force of 3200 N. a) Draw the FBD of the bin. b) Find the force of gravity acting on the bin. c) Find...
PHYSICS: Please help :) Thank you!
1. A crane pulls up on a bin with a mass of 250kg with a force of 3200 N.
a) Draw the FBD of the bin.
b) Find the force of gravity acting on the bin.
c) Find the acceleration of the bin.
2. Explain each of the following using the action and reaction forces:
a) A hockey player bounces back off the boards after running into them.
b) A runner can speed up.
3. Two students are standing on skateboards facing each other at rest. Student A (mass of 50.0 kg) pushes on student B (mass of 60.0 kg) with a force of 150 N [W]. You may assume there is no friction.
a) Draw the FBD of each skateboarder.
b) Find the acceleration of each skateboarder.
4. A helicopter of mass 4500 kg is hovering at rest above the ground.
a) Draw an FBD of the hovering helicopter.
b) Calculate the forces acting on the helicopter while it is hovering.
c) Explain what causes the upward force on the helicopter.
d) Find the acceleration of the helicopter if the propeller blades apply a force of 49 000 N [down] on the air.
1) When a crane pulls up on a bin, there are two forces acting on the bin: gravity, directed downward, and force from crane on the bin, directed up.
The force of gravity on the bin is `F_g = mg = 250*9.8 =2450 N` ` `
The net force on the bin is directed up, because the magnitude of the pull force of the crane, `F_(cr)` is greater than gravity. The magnitude of the net force F is
`F = F_(cr) - F_g = 3200 - 2450 =750 N`
The acceleration of the bin from the second Newton's Law will be
`a=F/m = 750/250 = 3 m/s^2`
2) Both of these statements can be explained using the third Newton's Law. It states that if there is a force acting from object A to object B, then there is an equal (in magnitude) and opposite (in direction) force acting from B to A. In other words, for every action there is a reaction.
If a hockey player runs into the boards, he pushes on them with some force (action.) Then, the boards push back with equal and opposite force (reaction) and the player bounces back off.
A runner can speed up because as (s)he runs, (s)he pushes on the ground in the direction opposite to running (action.) Therefore, the ground pushes back (reaction) and the runner speeds up.
3) The FBD of each skateboarder will include gravity (downward), normal force (upward) and "pushing" force: on student B, this force is directed West, and on student A, the force is equal and directed East. (See above why there is a force on student A.)
The acceleration of student B be will be to the West and equal
`a_B = F/m_B = 150/60 = 2.5 m/s^2`
The acceleration of student A will be to the East and equal
`a_A = F/m_A = 150/50 = 3 m/s^2`