A 150kg box is resting on a 150 incline with a coefficient of static friction of 0.35.What force acting along the plane is required to start the box moving up the plane?(878N)

Sure, here's how to solve the problem:

Step 1: Identify the forces acting on the box. These include the weight of the box (mass x gravity), the normal force (perpendicular to the incline), and the force of static friction.

Step 2: Calculate the weight of the box. This is the mass of the box times the acceleration due to gravity. In this case, it's 150 kg * 9.8 m/s^2 = 1470 N.

Step 3: Calculate the component of the weight force acting parallel to the incline. This is the weight times the sine of the angle of the incline. In this case, it's 1470 N * sin(15 degrees) = 382.68 N.

Step 4: Calculate the force of static friction. This is the coefficient of static friction times the normal force. The normal force in this case is the component of the weight force acting perpendicular to the incline, which is 1470 N * cos(15 degrees) = 1422.45 N. So, the force of static friction is 0.35 * 1422.45 N = 497.86 N.

Step 5: Calculate the force required to start the box moving up the plane. This is the force of static friction plus the component of the weight force acting parallel to the incline. So, it's 497.86 N + 382.68 N = 880.54 N.

So, approximately 881 N of force, acting along the plane, is required to start the box moving up the plane.