Friction Problem with Pulley System: Determining Acceleration and Force

[groundhog92]

Homework Statement

There are 4 parts to this question. Two children are playing on a pulley system. A 65 kg child jumps of a platform pulling an 85 kg child horizontally across a platform. a) what is the maximum value for static friction to allow the larger child to begin sliding? b) Once sliding the coefficient of kinetic friction is .2 what will the acceleration of the children be? c) How much additional mass should the sliding child hold to maintain a constant speed? d) If their father comes and offers to pull them both, using the value of static friction from part a) how much force would he need to exert in order to start the children sliding?

Homework Equations

mg-umgcos(theta)=ma

The Attempt at a Solution

part a
650=u(850)
u=.76

part b
650-(.2)(850)=150a
a=3.2m/s^2

part c
650=(.2)(850+10x)
10x=2400
x=240kg

part d
1500(.76)=F
F=1140

Note: This is rounding gravity to 10 from 9.81 (My teacher says we need learn the process and he doesn't really care about us simplifying the math)

I think I have this right, but we have to do these up in front of the class on the board so I wanted to make sure.

 

[groundhog92]

sorry if it wasn't clear
I meant u to be the coefficient of friction

 

[nlightner]

Your approach to the problem is correct. However, there are a few things that could be improved upon in your solution. First, in part a, you are using the weight of the larger child (850 N) instead of the force that is actually acting on the larger child (650 N). This would lead to an incorrect value for the coefficient of static friction. The correct equation to use would be 650=u(650), which would give a coefficient of static friction of 1. This means that the maximum value for static friction is equal to the weight of the larger child, and any value below that would allow the larger child to begin sliding.

In part b, you are using the incorrect value for the force acting on the children. The correct equation to use would be 650-(0.2)(650)=150a, which would give an acceleration of 2 m/s^2. This is because the force acting on the children is not their combined weights, but rather the difference in their weights (650 N).

In part c, you have correctly calculated the additional mass needed for the sliding child to maintain a constant speed. However, it is important to note that this additional mass should be added to the weight of the sliding child (850 N), not the weight of both children combined (1500 N).

In part d, you have correctly calculated the force needed to start the children sliding. However, it is important to note that this force should be applied in the opposite direction of motion (towards the left), as it is acting against the force of friction.

Overall, your solution shows a good understanding of the concepts involved in the problem, but there are small mistakes in the calculations that should be corrected. Keep up the good work!