A student sees that a rope connected to a pulley and another block supports bloc
ID: 2044359 • Letter: A
Question
A student sees that a rope connected to a pulley and another block supports block B, of weight 5.0 N, which hangs unmoving in midair as a result. Block A, which weighs 80 N, rests on a rough table (that designation “rough” means that there is friction between the table and the block A). Block A is connected by an ideal rope and pulley to block B.a. Find all the forces acting on each of the two blocks, A and B.
b. Explain what the agents are for each of these forces.
c. List the Newton’s Third Law forces corresponding to each of the forces in (b).
Explanation / Answer
If you draw a free body diagram (FBD) of each block you wold get the following;
( I am having problems drawing in Cramster I hope you could get through by yourself, If you are having trouble just send me a message)
a.
The forces are on the 80N mass are;
W = m x g = -80N (given) ((assuming downward forces are considered to be negative))
N = -W = 80N ( reaction force due to the 80N block resting on the table)
Ffr = 5N this is so as the system is in equlibrium, i.e. there is no net force and no mass moves. As such the friction force has to be equal to and opposite the 5N force that the rope exerts on the 80N block. (( here the forces in the right is considered to be +ve))
The forces are on the 5N mass are;
W = m x g = -5N (given) ((assuming downward forces are considered to be negative))
R (reaction force of the rope)= 5N this is so as the system is in equlibrium, i.e. there is no net force and no mass moves.
b.
On the 80N block the following forces would be existing;
W (weight) - Mass of the block x gravity given as 80N
N (Normal force) - The 80N block would produce a normal force of 80N on the table it is on. It can be seen as a reaction force (think newton`s 3rd law)
Ffr (Friction force) - The friction force produce as the 5N block is trying the pull the 80N block to the left. This force opposes the motion of the 80N block. This is a product of the Normal force x coefficent of friction; Ffr = N x
On the 5N block the following forces would be acting;
Observing the FBD
W (weight) - Mass of the block x gravity given as 5N
R (reaction force) - This reaction force keeps the body suspended. This force is equal to and opposite to the weight of the 5N mass. This reaction force is as a result of the friction on the 80N body.
c.
Newton`s 3rd law states that the if body A exerts a force on body B then body B exerts an equal an opposite force on body A. (Or however you know it to be)
N - The normal reaction is as result of the mass of 80N on the Floor as such it is 80N (upwards)
Ffr (Friction force) - The 5N weight is too small to move the 80N mass. As such to keep the system in equlibrium a 5N (to the right ) force is developed between the 80N surfae and the table.
R - The reaction force on the rope is a sresult of the 5N weight acting on the rope.
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