Solve numerically for the magnitude of the force Fm in Newtons.

(a) Fm = μs·N = μs·m·g(b) Fm = (0.61)(2.5)(9.8) = 14.95 N(c) F = Fm – μk·N = (μs – μk)N = (μs – μk)m·g = m·a ───► a = (μs – μk)g(d) a = (0.61 – 0.32)(9.8) = 2.84 m/s²

Ah..so you have started friction. So..do you know how frictional force works? Frictional forces can be classified into three broad types : Static , Dynamic and Rolling( Not required here, so we’ll skip it). Now, what exactly is friction? Its a force ( Resulting as a combination of the ragged surface on which the body moves and at times, due to the intermolecular forces of attraction between two highly smooth bodies) that hinders the motion of the moving body. When you try to move a static body, you will obviously have to apply some force on the body. Forces of friction will try to counter the force you provide. Say, you are applying some small force on a body. Frictional force will not the body to move, that is, it will adjust and provide just enough force to stop the motion of the body. This will continue untill you increase the force to a certain threshold after which the frictional force can no longer stop the body from motion. The body moves, however, there are still frictional forces that will reduce the acceleration of the body. Frictional force can be mathematically calculated as F=k*Normal Reaction. I am assuming that since you have started friction, you are aware of what normal reaction is. If not, Mail me. This k is called coefficient of friction. There are two kinds of coefficient of friction. One when the body is not moving. In that particular case the maximum force due to friction that will not allow the body to move is ks* Normal reaction, where ks is called co efficient of static friction. Once the body begins to move, however , the friction force is different The coefficient of friction becomes kd and the force is kd* Normal reaction where kd is called coefficient of kinetic friction. Curiously, kd is less than that of ks, meaning that if normal reaction is constant, the force of static friction is more than the force of kinetic friction, that is once the body begins to move you require less force to overcome friction. Now coming back to your problem. Clearly ks is 0.61 and kd is 0.32. Here the normal reaction is same as the weight of the body that’s 2.5g ( g is the acc due to gravity) So minimum force required to move the box is equal to the maximum force the static friction can offer, isn’t it? Once we cross this force the body will move. Now, as I have stated above, maximum force that static friction can offer is ks*Normal reaction of the box. Numerically it will be 2.5g*0.61 Newtons where g is roughly 9.8 m/s*s. This is FmOnce the box starts moving, we know its frictional force decreases. To how much? As I have stated earlier it becomes kd*Normal reaction. Numerically it will be 2.5g*0.32 Newtons, which is less than the force Fm we are applying right now. So there is a net force acting on the box that will help the box accelerate. How much is that? Our applied force Fm-Kinetic Frictional force, thats 2.5g(0.61 – 0.32)=2.5g*0.29 Newtons Remember that force =mass * acceleration. So net force acting on the body is 2.5g*0.29= 2.5 ( That’s the mass)* acceleration(a). Hence a=g*0.29. Assuming g=9.8, the value of a is 2.842 m/s*s. ( Many books may use the value of g to be 10 for simplicity. Others may use 9.81 for precision) Hope I was able to help. Sorry for the long reply, I wasn’t sure how much concept you previously had

You have :——————-W = box weight = ( m ) ( g ) = ( 2.5 kg ) ( 9.807 m/s^2 ) = 24.5175 N N = normal force = – W = – 24.5175 N Fm = (- N) ( mufs ) = ( m ) ( g ) ( mufs ) <——–[ a ]——Fm = ( 2.5 kg ) ( 9.807 m/s^2 ) ( 0.61 ) = 14.96 N —> <—–[ b ]——F = ( m ) ( g ) ( musf – mukf ) = ( m ) ( a )a = ( g ) ( musf – mukf ) <———–[ c ]—-a = ( 9.807 m/s^2 ) ( 0.61 – 0.32 ) = 2.84 m/s^2 <—-[ d ]——-

Please award Best Answer to the most helpful response to your question.