You pull with a force of255 N on a rope that is attached to a block of mass 30 kg, and the block slides across the floor at a constant speed of 1.1 m/s. The rope makes an angle $\mathit{\theta }\mathbf{=}{\mathbf{40}}^{\mathbf{\circ }}$with the horizontal. Both the force and the velocity of the block are in the xyplane.

(a) Express the tension force exerted by the rope on the block as a vector.

(b) Express the force exerted by the floor on the block as a vector.

The force acting on the given block is $F=255\mathrm{N}$

Mass of block is $m=30\mathrm{kg}$

Speed of the block is$v=1.1\mathrm{m}/\mathrm{s}$

Angle between rope and the horizontal is$\theta ={40}^{\circ }$

Figure 1 shows the direction of forces acting on the system.

Here, $f$is the frictional force acting on the block, mg is the weight of the block, $F$is the magnitude of the force acting on the block, $F_{\mathrm{N}}$is the normal force acting on the block and $v$is the magnitude of the velocity of the block.

Let's first provide a sketch of the problem in a shape of a free-body diagram of the block

Using equation we can express change of momentum as a net force:

$\frac{d\overrightarrow{p}}{dt}=\overrightarrow{F_{net}}$

Since the motion is at the constant speed, it means that the change of momentum is equal to 0 , which means that the net force is equal to 0.

Let's first observe direction x,

$$\begin{gathered} F_{net,x}=F·\cos \theta -T_x-F_f \\ =0 \\ {T}_x=F·\cos \theta \\ =255\mathrm{N}·\cos 40 \\ =195.34\mathrm{N} \end{gathered}$$

In the direction of y-axis, net force is a function of tension and the vertical component of the pull force:

$$\begin{gathered} F_{net,y}=F·\sin \theta -T_y \\ =0 \\ {T}_y=F·\sin \theta \\ =255\mathrm{N}·\sin 40 \\ =163.91\mathrm{N} \end{gathered}$$

Now that we have determined magnitude of the vertical and horizontal component of the tension force, we can write it down as a vector:

$\overrightarrow{T}=\left(195.34\mathrm{N},163.91\mathrm{N},0\right)$

In this case, net force is found as a discrepancy between the normal force and the weight of the block. If the net force is to be 0, we have the following expression then:

$$\begin{gathered} \frac{d\overrightarrow{p}}{dt}={\overrightarrow{F}}_{net}=0 \\ {F}_{net}=F_N-W \\ {F}_N-W=0 \\ {F}_N=W \end{gathered}$$

In correspondence with mass of the block and the gravitational acceleration, we will determine the weight of the block, which is equal to the normal force by the floor:

$$\begin{gathered} F_N=W \\ =mg \\ =30\mathrm{kg}\times 9.81\mathrm{m}/{\mathrm{s}}^2 \\ =294.3\mathrm{N} \end{gathered}$$

Since there is no $x$nor $z$component of force $F_N$, this is the force shown in vector form:

${\overrightarrow{F}}_N=\left(0,294.3,0\right)\mathrm{N}$

The net force is zero, so, let us solve the equation for $F_f$and plug the values for $F_x$

$$\begin{gathered} F_f=F_{net,x}-F_x \\ =0-350 \\ =-350\mathrm{N} \end{gathered}$$

Calculate the x-component of the floor force $F_{f,x}=-350\mathrm{N}$ Also, there is no force component in z-direction or in y direction by the floor, therefore, $F_f,y=0$and $F_f,z=0$and the force vector in the three directions is given by,

${\overrightarrow{F}}_f=F_{f,x}\hat{i}+F_{f,y}\hat{j}+F_{f,z}\hat{k}$

Now we can plug our values for $F_{f,x},F_{f,y}$and $F_{f,z}$into above equation to get the vector of the floor force

$$\begin{gathered} {\overrightarrow{F}}_f=F_{f,x}\hat{i}+F_{f,y}\hat{j}+F_{f,z}\hat{k} \\ =\left(-350\hat{i}+0\hat{j}+0\hat{k}\right) \\ =\left(-350\hat{i}\right)\mathrm{N} \end{gathered}$$

Therefore, the force exerted by the floor on the block as a vector is ${\overrightarrow{F}}_f=\left(-350\hat{i}\right)\mathrm{N}$.