Question

1. two forces are applied to a car in an effort to move it, as shown in the picture. what is the resultant of these two forces? what direction is the force relative to mathematical 0° (assume the car is pointing towards mathematical 0°)? if the car remains stationary, what is the magnitude and direction of the friction produced by the wheels?

Explanation:

Step1: Resolve forces into x and y components

Let's define the positive x - axis as the direction of \(0^{\circ}\) (the direction the car is pointing). Let \(F_1 = 450\space N\) with an angle \(\theta_1=10^{\circ}\) from the x - axis and \(F_2 = 400\space N\) with an angle \(\theta_2 = 10^{\circ}+ 30^{\circ}=40^{\circ}\) from the x - axis.

The x - component of a force \(F\) with angle \(\theta\) from the x - axis is given by \(F_x=F\cos\theta\) and the y - component is given by \(F_y = F\sin\theta\).

For \(F_1\):
\(F_{1x}=450\cos(10^{\circ})\approx450\times0.9848 = 443.16\space N\)
\(F_{1y}=450\sin(10^{\circ})\approx450\times0.1736 = 78.12\space N\)

For \(F_2\):
\(F_{2x}=400\cos(40^{\circ})\approx400\times0.7660 = 306.4\space N\)
\(F_{2y}=400\sin(40^{\circ})\approx400\times0.6428 = 257.12\space N\)

Step2: Calculate the resultant x and y components

The resultant x - component \(R_x=F_{1x}+F_{2x}\)
\(R_x = 443.16+306.4=749.56\space N\)

The resultant y - component \(R_y=F_{1y}-F_{2y}\) (assuming \(F_{1y}\) and \(F_{2y}\) are in opposite y - directions, we take the difference. If the diagram shows both forces below the dashed line, maybe \(F_{1y}\) and \(F_{2y}\) are in the same direction, but from the angle notation, let's assume \(F_1\) is \(10^{\circ}\) to the right of the dashed line (x - axis) and \(F_2\) is \(30^{\circ}\) to the left of \(F_1\), so \(F_{1y}\) and \(F_{2y}\) are in opposite y - directions. Let's correct: If the dashed line is the x - axis, and \(F_1\) is \(10^{\circ}\) above the x - axis (north - east component) and \(F_2\) is \(30^{\circ}\) below the x - axis (south - east component), then \(F_{1y}\) is positive (north) and \(F_{2y}\) is negative (south). So \(R_y=F_{1y}+F_{2y}\) (if \(F_2\) is below, \(F_{2y}\) is negative). Wait, maybe the angles are measured from the x - axis. Let's re - interpret: Let the x - axis be the direction the car is pointing (0°). The first force \(F_1 = 450\space N\) is at \(10^{\circ}\) above the x - axis, the second force \(F_2=400\space N\) is at \(10^{\circ}+ 30^{\circ}=40^{\circ}\) below the x - axis. So:

\(F_{1x}=450\cos(10^{\circ})\), \(F_{1y}=450\sin(10^{\circ})\)

\(F_{2x}=400\cos(40^{\circ})\), \(F_{2y}=- 400\sin(40^{\circ})\) (negative because it's below the x - axis)

Then \(R_y=F_{1y}+F_{2y}=450\sin(10^{\circ})-400\sin(40^{\circ})\)

\(450\sin(10^{\circ})\approx450\times0.1736 = 78.12\)

\(400\sin(40^{\circ})\approx400\times0.6428 = 257.12\)

\(R_y=78.12 - 257.12=- 179\space N\) (negative means south direction)

\(R_x=450\cos(10^{\circ})+400\cos(40^{\circ})\approx450\times0.9848 + 400\times0.7660\)

\(450\times0.9848 = 443.16\), \(400\times0.7660 = 306.4\)

\(R_x=443.16 + 306.4=749.56\space N\)

Step3: Calculate the magnitude of the resultant force

The magnitude of the resultant force \(R\) is given by \(R=\sqrt{R_x^{2}+R_y^{2}}\)

\(R=\sqrt{(749.56)^{2}+(- 179)^{2}}=\sqrt{749.56^{2}+179^{2}}\)

\(749.56^{2}\approx749.56\times749.56\approx561840\)

\(179^{2}=32041\)

\(R=\sqrt{561840 + 32041}=\sqrt{593881}\approx770.6\space N\)

Step4: Calculate the direction of the resultant force

The direction \(\theta\) of the resultant force with respect to the x - axis is given by \(\tan\theta=\frac{\vert R_y\vert}{R_x}\) (since \(R_y\) is negative, the angle is below the x - axis)

\(\tan\theta=\frac{179}{749.56}\approx0.239\)

\(\theta=\arctan(0.239)\approx13.4^{\circ}\) below the x - axis.

Step5: Friction force (if the car is stationary)

If the car remains stationary, the friction force \(f\) is equal in magnitude and opposite in direction to the resultant force of the two applied forces. So the magnitude of the friction force is equal to the magnitude of the resultant force of the two applied forces, and the direction is opposite to the resultant force of the two applied forces.

Answer:

The magnitude of the resultant force of the two applied forces is approximately \(771\space N\) (direction approximately \(13.4^{\circ}\) below the direction the car is pointing). The magnitude of the friction force is also approximately \(771\space N\) and its direction is opposite to the resultant force of the two applied forces.