Question

A box of mass m = 19.5 kg m=19.5 kg is pulled up a ramp that is inclined at an angle θ = 23.0 ∘ θ=23.0∘ angle with respect to the horizontal. The coefficient of kinetic friction between the box and the ramp is μ k = 0.335 μk=0.335 , and the rope pulling the box is parallel to the ramp. If the box accelerates up the ramp at a rate of a = 2.69 m/s 2 a=2.69 m/s2 , calculate the tension F T FT in the rope. Use g = 9.81 m/s 2 g=9.81 m/s2 for the acceleration due to gravity.

Answer 1

Answer:

186 N

Explanation:

Mass of box=m=19.5 kg

$\theta=23^{\circ}$

Coefficient of friction between the box and ramp=$\mu_k=0.335$

$a=2.69 m/s^2$

Acceleration due to gravity,g=$9.81 m/s^2$

$T-mgsin\theta-\mu_kmgcos\theta=ma$

Substitute the values

$T-19.5\times 9.81sin23-0.335\times 19.5\times 9.81cos23=19.5\times 2.69$

$T=19.5\times 9.81sin23+0.335\times 19.5\times 9.81cos23+19.5\times 2.69$

$T=186 N$

Answer 2

Answer:

The tension in the rope is 186.0 N.

Explanation:

Given that,

Mass of box = 19.5 kg

Angle = 23.0°

Coefficient of kinetic friction = 0.335

Acceleration = 2.69 m/s²

In y direction, the acceleration is zero.

So. The net force is zero.

We need to calculate the normal force

Using balance equation

$N-mg\cos\theta=ma$

Where, n = normal force

a = acceleration

Put the value into the formula

$N-19.5\times9.8\cos23=0$

$N=175.90\ N$

We need to calculate the friction force

Using formula of friction force

$F_{\mu k}=\mu N$

Put the value into the formula

$F_{\mu k}=0.335\times175.90$

$F_{\mu k}=58.92\ N$

We need to calculate the tension

Using balance equation

$T-mg\sin\theta-F_{\mu k}=ma$

$T=ma+mg\sin\theta+F_{\mu k}$

$T=19.5\times2.69+19.5\times9.8\sin23+58.92$

$T=186.0\ N$

Hence, The tension in the rope is 186.0 N.