- #1
subzero0137
- 91
- 4
The problem as it appears on the problem sheet: A block of weight [itex]mg = 500 N[/itex] is sitting on a horizontal floor. The coefficient of kinetic friction is [itex]\mu_{k} = 0.4[/itex]. In class, we determined at what angle from the horizontal I should pull on the block if I'm weak and need to minimise the force [itex]F[/itex] with which I am pulling. Since then, I've been working out, and now I'm very strong - but also very lazy. At what angle [itex]\theta[/itex] from the horizontal should I pull the block in order to minimise the total work I need to do in order to move the block by [itex]10 m[/itex] in a straight horizontal line at a constant, non-zero velocity, and what will that work be?
Relevant equations:
(1)$$N+F\sin(\theta)= 500$$
(2)$$F\cos(\theta)=friction=0.4N$$
(3)$$N=\frac{F\cos(\theta)}{0.4}$$
Plugging (3) in (1) gives: $$\frac{F\cos(\theta)}{0.4}+F\sin(\theta)=500$$ which gives force [itex]F[/itex] as a function of [itex]\theta[/itex]:
(4)$$F=\frac{500}{\frac{\cos(\theta)}{0.4}+\sin(\theta)}$$
(5)$$W=F\cos(\theta)\times10$$
In a previous lecture, the teacher derived (4), and showed that to minimise [itex]F[/itex], the denominator in (4) has to maximised. So he took the denominator, differentiated it with respect to [itex]\theta[/itex], and set the derivative to equal to 0 (to determine the maxima):
(4.5)$$\cos(\theta)-\frac{\sin(\theta)}{0.4} = 0 ∴ \theta = \arctan(0.4) = 21.8^{o}$$
My attempt: (5) is the definition of work done, and I need to find the angle at which [itex]W[/itex] is the minimum. Clearly, [itex]W[/itex] is a multivariable function since not only does [itex]10\cos(\theta)[/itex] vary as [itex]\theta[/itex] varies, but [itex]F[/itex] also varies since it is also a function of [itex]\theta[/itex]. So substituting (4) in (5) gives (6): $$W=\frac{500}{\frac{\cos(\theta)}{0.4}+\sin(\theta)}\times 10\cos(\theta) = \frac{5000}{2.5+\tan(\theta)}$$
For [itex]W[/itex] to be minimised, the denominator of (6): [itex]2.5+\tan(\theta)[/itex] must be maximised, so I did what my teacher did, and simply differentiated the denominator and set the derivative to equal to 0: $$\sec^{2}(\theta) = 0$$ However, this does not give me a physically meaningful value for the angle. Where did I go wrong, and what do I do? I've considered various alternative forms of (5), but none seem to give any real solutions for the angle.
Any help would be appreciated.
Relevant equations:
(1)$$N+F\sin(\theta)= 500$$
(2)$$F\cos(\theta)=friction=0.4N$$
(3)$$N=\frac{F\cos(\theta)}{0.4}$$
Plugging (3) in (1) gives: $$\frac{F\cos(\theta)}{0.4}+F\sin(\theta)=500$$ which gives force [itex]F[/itex] as a function of [itex]\theta[/itex]:
(4)$$F=\frac{500}{\frac{\cos(\theta)}{0.4}+\sin(\theta)}$$
(5)$$W=F\cos(\theta)\times10$$
In a previous lecture, the teacher derived (4), and showed that to minimise [itex]F[/itex], the denominator in (4) has to maximised. So he took the denominator, differentiated it with respect to [itex]\theta[/itex], and set the derivative to equal to 0 (to determine the maxima):
(4.5)$$\cos(\theta)-\frac{\sin(\theta)}{0.4} = 0 ∴ \theta = \arctan(0.4) = 21.8^{o}$$
My attempt: (5) is the definition of work done, and I need to find the angle at which [itex]W[/itex] is the minimum. Clearly, [itex]W[/itex] is a multivariable function since not only does [itex]10\cos(\theta)[/itex] vary as [itex]\theta[/itex] varies, but [itex]F[/itex] also varies since it is also a function of [itex]\theta[/itex]. So substituting (4) in (5) gives (6): $$W=\frac{500}{\frac{\cos(\theta)}{0.4}+\sin(\theta)}\times 10\cos(\theta) = \frac{5000}{2.5+\tan(\theta)}$$
For [itex]W[/itex] to be minimised, the denominator of (6): [itex]2.5+\tan(\theta)[/itex] must be maximised, so I did what my teacher did, and simply differentiated the denominator and set the derivative to equal to 0: $$\sec^{2}(\theta) = 0$$ However, this does not give me a physically meaningful value for the angle. Where did I go wrong, and what do I do? I've considered various alternative forms of (5), but none seem to give any real solutions for the angle.
Any help would be appreciated.