Answer to Question #197106 in Mechanics | Relativity for Victor Oyejide

Question

A rope inclined at angle 37degree to the horizontal is used to drag a 50kg block along a level floor with an acceleration of 1m/s². The coefficient of friction between the block and floor is 0.2. What is the tension in the rope?

Answer 1

Gives

$\theta=37°$

$a=1m/sec$

m=50kg

$\mu=.2$

We know that

Horizontal components

$f=Tcos37°$

Vertical components

Tsin37°+N=mg

N=mg-Tsin37°

We can rewrite horizontal components

f=Tcos37°

$\mu N=Tcos37°$

Put N value

$\mu(mg-Tsin37°)=Tcos30°$

$\mu mg=T(\mu sin30°+cos30°)$

$T=\frac{\mu mg}{\mu sin37°+cos37°}$

Put value

$T=\frac{0.2\times50\times9.8}{0.2sin37°+cos37°}$

$T=\frac{98}{0.9189}=106.64$ N