Question

A worker drags a crate across a factory floor by pulling on a rope tied to the crate. The worker exerts a force of magnitude F 450 N on the rope, which is inclined at an upward angle (theta) = 38° to the horizontal, and the floor exerts a horizontal force of magnitude f = 125 N that opposes the motion. Calculate the magnitude of the acceleration of the crate if (a) its mass is 310 kg and (b) its weight is 310 N.

Answer 1

Answer:

a) a= 0.74 m/s²

b) a=7.25 m/s²

Explanation:

We apply Newton's second law to the crate:

∑F = m*a (Formula 1)

∑F : algebraic sum of the forces in Newton that are acting on the crate (N)

m : mass of the crate  (kg)

a : acceleration on the crate  (m/s²)

Look at the free body diagram of the crate in the attached graphic

F =450 N :force of  the rope, which is inclined at an upward angle θ = 38° to the horizontal

x-y  components of F

Fx =450*cos38°= 354.6 N

Fx =450*sin38° = 277.05 N

force of magnitude f

f = 125 N  : in the opposite direction to the movement of the crate

a)Calculation of the  acceleration of the crate if its mass is 310 kg

We apply the formula (1) in direction x:

∑F = m*a

Fx-f =  310*a

$a=\frac{354.6-125}{310}$

a= 0.74 m/s²

a)Calculation of the  acceleration of the crate if its mass is 310 kg

W= m*g  m= W/g  

m= W/g  = 310/9.8= 31.63 Kg

We apply the formula (1) in direction x:

∑F = m*a

Fx-f =  31.63*a

$a=\frac{354.6-125}{31.63}$

a=7.25 m/s²