A force of magnitude 50 N is acting at an angle of 30° on a 2 kg object to push the object against a rough horizontal ceiling as shown in the diagram below. The object accelerates to the LEFT at 20 m.s-2. Ignore air resistance.

The object is not accelerating vertically, so the net force in the vertical direction is zero ($\sum F_y = 0$).
The forces acting vertically are:

• Weight ($F_g$) acting downwards.
• Normal force ($F_N$) from the ceiling, acting downwards (as the object is pushed against the ceiling from below).
• Vertical component of the applied force ($F_y$) acting upwards.

$$F_g = mg$$ $$F_g = (2 \text{ kg})(9.8 \text{ m/s}^2)$$ $$F_g = 19.6 \text{ N}$$

$$F_y = F \sin(\theta)$$ $$F_y = (50 \text{ N}) \sin(30^\circ)$$ $$F_y = (50 \text{ N})(0.5)$$ $$F_y = 25 \text{ N}$$

Taking upwards as positive:

$$\sum F_y = F_y - F_g - F_N = 0$$ $$25 \text{ N} - 19.6 \text{ N} - F_N = 0$$ $$5.4 \text{ N} - F_N = 0$$ $$F_N = 5.4 \text{ N}$$

The magnitude of the normal force is $\boxed{\text{5.4 N}}$.

2.3.2 Magnitude of the coefficient of kinetic friction.

The object accelerates to the left, so the net force in the horizontal direction is $\sum F_x = ma_x$.
The forces acting horizontally are:

• Horizontal component of the applied force ($F_x$) acting to the left.
• Kinetic friction ($F_k$) acting to the right (opposing the leftward motion).

$$F_x = F \cos(\theta)$$ $$F_x = (50 \text{ N}) \cos(30^\circ)$$ $$F_x = (50 \text{ N})\left(\frac{\sqrt{3}}{2}\right)$$ $$F_x \approx 43.30 \text{ N}$$

Taking leftwards as positive:

$$\sum F_x = F_x - F_k = ma_x$$

We know $F_k = \mu_k F_N$.

$$F_x - \mu_k F_N = ma_x$$ $$(43.30 \text{ N}) - \mu_k (5.4 \text{ N}) = (2 \text{ kg})(20 \text{ m/s}^2)$$ $$43.30 \text{ N} - 5.4 \mu_k = 40 \text{ N}$$

$$43.30 - 40 = 5.4 \mu_k$$ $$3.30 = 5.4 \mu_k$$ $$\mu_k = \frac{3.30}{5.4}$$ $$\mu_k \approx 0.611$$

The magnitude of the coefficient of kinetic friction is $\boxed{\text{0.61}}$.

2.4 The force is now acting at a smaller angle. What effect will this have on the following? Choose from INCREASES, DECREASES or REMAINS THE SAME.

2.4.1 Value calculated in QUESTION 2.3.2

The value calculated in QUESTION 2.3.2 is the coefficient of kinetic friction ($\mu_k$). The coefficient of kinetic friction depends only on the nature of the surfaces in contact, not on the angle or magnitude of the applied force.

The value calculated in QUESTION 2.3.2 REMAINS THE SAME.

2.4.2 Kinetic frictional force acting on the object. Give a reason for the answer.

The kinetic frictional force is given by $F_k = \mu_k F_N$.
From 2.3.1, the normal force is $F_N = F \sin(\theta) - F_g$.
If the angle $\theta$ (above the horizontal) becomes smaller, then $\sin(\theta)$ will decrease.
Since $F_N = F \sin(\theta) - F_g$, a decrease in $F \sin(\theta)$ will cause $F_N$ to decrease.
As $\mu_k$ remains the same (from 2.4.1) and $F_N$ decreases, the kinetic frictional force $F_k = \mu_k F_N$ will DECREASE.

Reason: A smaller angle of the applied force (above the horizontal) means a smaller upward vertical component of the applied force. This results in a smaller normal force exerted by the ceiling on the object, which in turn leads to a smaller kinetic frictional force.