Question

A 20 kg wooden crate is dragged 12 m across a rough level floor by a rope. The force of tension in the rope is 50 N at an angle of 25° above the horizontal and the crate is moving at a constant speed of 1.50 m/s.

1. Find the horizontal component of the force of tension in [N].
2. Find the vertical component of the force of tension in [N].
3. Find the magnitude of the weight of the crate in [N].
4. Find the magnitude of the normal force acting on the crate in [N].
5. Find the magnitude of the friction force acting on the crate in [N].
6. Find the work done by the tension force in [J].
7. Find the work done by weight in [J].
8. Find the work done by the normal force in [J].
9. Find the magnitude of the work done by the friction force in [J].
10. What is the total work done on the object in [J]?
11. Find the coefficient of the crate on the floor.

Answer 1

Explanations & Calculations

1)

• You need to know how to resolve a given force with respect to the direction needed.
• For this case, the horizontal component of the force will be

$\qquad\qquad \begin{aligned} \small F_h&=\small 50N\times\cos25\\ &=\small \bold{45.32\,N} \end{aligned}$

2)

• And the verticle component will be

$\qquad\qquad \begin{aligned} \small F_v &=\small 50N\times\sin25\\ &=\small \bold{21.13\,N} \end{aligned}$

3)

• The weight of any object is obtained by multiplying its own mass with the gravitational acceleration: g. Therefore, weight of this crate will be

$\qquad\qquad \begin{aligned} \small w&=\small mg=20kg\times9.8ms^{-2}\\ &=\small \bold{196\,N} \end{aligned}$

4)

• To access the normal force, we need to consider the equilibrium of forces along the verticle direction. Since the crate is not moving along that direction, all the forces along that direction need to be in equilibrium. Therefore,

$\qquad\qquad \begin{aligned} \small \uparrow\Sigma F &=\small 0\\ \small R+T\sin25-w&=\small 0\\ \small R&=\small w-F_v=196-21.13\\ &=\small \bold{174.87\,N} \end{aligned}$

5)

• For this, we need to access the behavior of forces along the moving direction as the friction acts parallelly along that direction. Since the crate is moving at a constant speed, no acceleration is present along that direction hence all the forces are perfectly balanced out. Therefore,

$\qquad\qquad \begin{aligned} \small \to \Sigma F&=\small 0\\ \small T\cos25-f&=\small 0\\ \small f&=\small T\cos25=F_h\\ &=\small \bold{45.32\,N} \end{aligned}$

6)

• Since the thread is positioned in a slanted direction, only its horizontal component performs work as it is the only component that can move with time. Therefore,

$\qquad\qquad \begin{aligned} \small W_T&=\small F_h\times s =45.32N\times12m\\ &=\small \bold{543.84\,J} \end{aligned}$

7)

• Since the line of action of weight does not move with time: it remains directed down, the work done by it is zero.

$\qquad\qquad \begin{aligned} \small W_w&=\small mg\times s=mg\times0\\ &=\small \bold{0\,J} \end{aligned}$

8)

• The same as it was for the weight, its 0.

9)

• Work done by the friction will be,

$\qquad\qquad \begin{aligned} \small W_f&=\small f\times s = 45.32N \times 12m\\ &=\small \bold{543.84\,N} \end{aligned}$

10)

• The total work is what is done by the thread as it is the doer source in this situation. All the work it does is what we calculated above,

$\qquad\qquad \begin{aligned} \small W_{total}&=\small W_h\\ &=\small \bold{543.84\,J} \end{aligned}$

11)

• By considering the equation for static friction, the coefficient can be assessed.

$\qquad\qquad \begin{aligned} \small F&=\small \mu R\\ \small \mu &=\small \frac{f}{R}= \frac{45.32N}{174.87N}\\ &=\small \bold{0.26} \end{aligned}$

• What happens here is that the tension provides just the energy\work needed to overcome the friction & maintain that speed constantly over the motion: see how the total work equals the work done by the friction.