A 160 N block rest on a table attached to a string and pulley. The suspended mass over the side of the table has a weight of 53 N.
PART 1: What is the magnitude of the minimum force of static friction required to hold both blocks at rest? Answer in units of N.
PART 2: What minimum coefficient of static friction is required to ensure that both blocks remain at rest?
Elizabeth,
The tutors are not going to take the exams for you at the end of the term. This is your chance to attempt some questions on friction, test your understanding and make mistakes if you have to.
Post your attempts, and if there are mistakes, we will be glad to help you.
It is pointless to submit perfect homework but get a 25% at the exam.
i have done all my homework these 5 problems i could not understand.. my answers are due at midnight tonight im a senior in this physics course and math is generally my strong point but when it comes to things like this i try and i didn't get it so i posted the problem on here looking for help... sry if 5 problems out of 32 is such a huge problem to get help with...
Bbc
To find the magnitude of the minimum force of static friction required to hold both blocks at rest, we need to consider the forces acting on the system.
PART 1:
1. The weight of the 160 N block acts vertically downward with a force of 160 N.
2. The weight of the suspended mass acts vertically downward with a force of 53 N.
3. The tension in the string attached to the pulley pulls the 160 N block horizontally with a force of 160 N (assuming an ideal and massless string).
4. The static friction force holds the 160 N block in place horizontally on the table.
Since the system is at rest, the forces in the horizontal direction are balanced. Therefore, the static friction force must equal the tension in the string.
So, the magnitude of the minimum force of static friction required to hold both blocks at rest is 160 N.
PART 2:
To find the minimum coefficient of static friction required to ensure that both blocks remain at rest, we can use the equation: μs = Fs / N, where μs is the coefficient of static friction, Fs is the force of static friction, and N is the normal force.
In this case, the normal force N is equal to the weight of the 160 N block, which is also 160 N. Therefore, we have:
μs = Fs / N
= 160 N / 160 N
= 1
So, the minimum coefficient of static friction required to ensure that both blocks remain at rest is 1.