3 - A box of mass 2 Kg lies on a rough horizontal floor with the coefficient of friction between the floor and the box being 0.5 (Figure 3). A light string is attached to the box in order to pull the box across the floor. If the tension in the string is TN, find the value that T must exceed for motion to occur if the string is 30º above the horizontal.


Figure 3

Determine the distance to which the 90 kg painter can climb without causing the 4 m ladder to slip at its lower end A as shown in Figure 1. The top of the 15 kg ladder has a smaller roller, and at the ground the coefficient of static friction is 0.25. The mass center of the painter is directly above his feet.


Figure 1

232

To solve this problem, we need to consider the forces acting on the box and determine the minimum tension in the string required for motion to occur.

First, let's analyze the forces acting on the box. We have the weight force (mg) acting vertically downward, where m is the mass of the box (2 kg) and g is the acceleration due to gravity (9.8 m/s²). We also have the normal force (N) acting vertically upward to balance the weight force.

Since the box is on a rough horizontal surface, there is also a frictional force (f) between the box and the floor, which opposes the direction of motion. The magnitude of the frictional force can be calculated using the equation f = µN, where µ is the coefficient of friction (0.5) and N is the normal force.

Next, let's consider the tension force (T) in the string. The tension force can be divided into two components: one component acting vertically upward (Tcosθ) and one component acting horizontally in the direction of motion (Tsinθ), where θ is the angle the string makes with the horizontal (30º in this case).

For the box to start moving, the horizontal component of the tension force (Tsinθ) must be greater than or equal to the frictional force (f). Mathematically, we have Tsinθ ≥ f.

Substituting the expressions for f and Tsinθ, we have Tsinθ ≥ µN.

Since the normal force (N) is equal to the weight force (mg) in this case, we can further simplify the expression to Tsinθ ≥ µmg.

Finally, we can solve for the minimum tension force (T) by rearranging the inequality Tsinθ ≥ µmg. Dividing both sides by sinθ, we get T ≥ µmg/sinθ.

Plugging in the given values, T ≥ (0.5)(2 kg)(9.8 m/s²)/sin(30º).

Now, we can calculate the minimum tension force required for motion to occur.