Aworker 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 θ = 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

(a) a= 0.74 m/s²

(b) a= 7.26 m/s²

Explanation:

Known data

a)m= 310 kg , : crate mass

F = 450 N which is inclined at an upward angle θ = 38° to the horizontal, in direction of the motion

f= 125 N opposes the motion.

b)W = 310 N  :crate weight

F = 450 N which is inclined at an upward angle θ = 38° to the horizontal, in direction of the motion

f= 125 N opposes the motion

F component in x

Fx= F*Cosθ = 450*Cos38°

Fx= 354.6 N

F component in y

Fy= F*sinθ = 450*sin38°

Fy= 277.05N

Newton's second law:

∑F = m*a (Formula 1)

∑F : algebraic sum of the forces in Newton (N)

m : mass in kilograms (kg)

a : acceleration in meters over second square (m/s²)

Problem development

We choose the positive direction, the direction of the crate's movement and apply equation 1:

a)∑Fx = m*a

Fx-f= m*a

354.6 - 125 = 310*a

310*a = 229.6

a= 0.74 m/s²

b) ∑Fx = m*a  ,   m= W/g  = 310/9.8 =31.63 N

Fx-f= m*a

354.6 - 125 = 31.63*a    

31.63*a  = 229.6

a= 7.26 m/s²

true tan speed = ang vel x radius