A machine in an ice factory is capable of exerting 3.31 × 102 N of force to pull large blocks of ice up a slope. The blocks each weigh 1.32 × 104 N. Assuming there is no friction, what is the maximum angle that the slope can make with the horizontal if the machine is to be able to complete the task? Answer in units of ◦ .

Question

contestada

Respuesta :

ACCESS MORE ACCESS MORE ACCESS MORE ACCESS MORE

nuuk nuuk · Answer 1 · 2019-11-19T07:54:09+01:00

Answer:[tex]\theta =1.43 ^{\circ}[/tex]

Explanation:

Given

Force exerted by machine [tex]F=3.31 \times 10^2 N[/tex]

Weight of Block [tex]W=mg=1.32\times 10^4 N[/tex]

Considering inclination to be [tex]\theta [/tex]

After resolving Forces We get

[tex]mg\sin \theta =F[/tex]

[tex]\sin \theta =\frac{F}{mg}[/tex]

[tex]\sin \theta =\frac{3.31 \times 10^2}{1.32\times 10^4}[/tex]

[tex]\sin \theta =2.507\times 10^{-2}[/tex]

[tex]\sin \theta =0.02507[/tex]

[tex]\theta =\sin ^{-1}(0.02507)[/tex]

[tex]\theta =1.43 ^{\circ}[/tex]

fichoh fichoh · Answer 2 · 2021-10-23T11:37:04+02:00

The maximum angle which the slope can make with the horizontal if the task is to be completed is [tex] 1.44° [/tex]

Given the Parameters :

The force made with the horizontal, we use the relation :

Making [tex] \theta [/tex] the subject :

[tex] \theta = sin^{-1}[ \frac{F}{Mg} ][/tex]

[tex] \theta = sin^{-1}[ \frac{3.31 \times 10^{2}}{1.32 \times 10^{4} } ][/tex]

[tex] \theta = sin^{-1}[ 0.0250757 ] = 1.436 [/tex]

Therefore, the maximum angle the slope can make with the horizontal in this scenario is [tex] 1.44° [/tex]

Learn more :https://brainly.com/question/14897423