The co-efficient for static friction for a 5 kg box on a floor is .72. How much force if required to start the box sliding?
I worked it as follows:
Force of friction=Force normal * co-effic. of friction which is .72*5*9.8=35.28 but answer in back of my book says this is wrong. Where did I err?
Force of Friction= Force of the normal*co-effic...
Coefficient= .72
Normal Force= mass*g
9.81*5
49.05
Solve: 49.05*0.72
35.316 or 35.32
Is that what you got? If it is, you're right. If not, recheck the book, it may be that you are looking at the wrong answer.
Yes, that's my answer too. Incorrect answers at the back of the book drive me crazy! thanks for confirming!
From your calculations, it seems that you used the correct formula for calculating the force of friction. However, there might be a small mistake in your calculations. Let's go through the calculation step by step:
1. Calculate the force normal:
The weight of the box can be calculated using the formula weight = mass * acceleration due to gravity. In this case, the mass is 5 kg and the acceleration due to gravity is approximately 9.8 m/s^2. Therefore, the weight of the box is:
Weight = 5 kg * 9.8 m/s^2 = 49 N.
2. Calculate the force of friction:
The formula for the force of friction is force of friction = force normal * coefficient of static friction. In this case, the coefficient of static friction is given as 0.72 and the force normal (weight) is 49 N. Therefore, the force of friction is:
Force of friction = 49 N * 0.72 = 35.28 N.
Based on your calculations, it seems like you have calculated the force of friction correctly.
If the answer in the back of your book says this is wrong, there might be an error in the given coefficient of static friction or another factor that you need to consider. Double-check the equation provided in the book and make sure you have not missed any additional information or steps in the problem.