You push downward on a box at an angle 25∘ below the horizontal with a force of 750 N. If the box is on a flat horizontal surface for which the coefficient of static friction with the box is 0.65, what is the mass of the heaviest box you will be able to move? Balancing of forces:When two or more forces act on a body and the body is in equilibrium, the forces are balanced in horizontal and vertical directions respectively. The component of a force along a direction making an angle θ with the direction is: Fcosθ and the componentperpendicular to that direction is Fsinθ.

74.29 kg

Step-by-step explanation:

Force applied = 750 N, coefficient of static friction = 0.65

the force applied has vertical and horizontal component

horizontal component (force tending to move the body) = 750 cos 25° = 679.731 N

vertical component of the force = 750 sin 25° = 316.96

coefficient of static friction = frictional force / force of normal

force of normal acting upward = weight of the box + the vertical component of the force acting downward. = mg + 316.96

0.65 = Max frictional force / (mg + 316.96) where frictional = 679.731 since in the opposite direction

0.65 = 679.731 / (mg + 316.96)

mg + 316.96 = 679.731 / 0.65 = 1045.74

mg = 1045.74 - 316.96 = 728.78

where g is 9.81 m/s² acceleration due to gravity

9.81 m = 728.78

m = 728.78 / 9.81 = 74.29 kg