Ask question

Ask question

All categories

1 year ago

5

A man is dragging a trunk up the loading ramp of a mover’s truck. The ramp has a slope angle of 20.0°, and the man pulls upward

with a force FS whose direction makes an angle of 30.0° with the ramp (Fig. E4.4). (a) How large a force FS is necessary for the component Fx parallel to the ramp to be 90.0 N?(b) How large will the component Fy perpendicular to the ramp be then?

1 answer:

AleksandrR [38]1 year ago

5 0

Answer:

(a) 104 N

(b) 52 N

Explanation:

Given Data

Angle of inclination of the ramp: 20°

F makes an angle of 30° with the ramp

The component of F parallel to the ramp is Fx = 90 N.

The component of F perpendicular to the ramp is Fy.

(a)

Let the +x-direction be up the incline and the +y-direction by the perpendicular to the surface of the incline.

Resolve F into its x-component from Pythagorean theorem:

Fx=Fcos30°

Solve for F:

F= Fx/cos30°

Substitute for Fx from given data:

Fx=90 N/cos30°

=104 N

(b) Resolve r into its y-component from Pythagorean theorem:

Fy = Fsin 30°

Substitute for F from part (a):

Fy = (104 N) (sin 30°)

= 52 N

You might be interested in

A resultant vector is 8.00 units long and makes an angle of 43.0 degrees measured �� – �� with respect to the positi

Komok [63]

Answer:

223 degree

Explanation:

We are given that

Magnitude of resultant vector= 8 units

Resultant vector makes an angle with positive -x in counter clockwise direction

$\theta=43^{\circ}$

We have to find the magnitude and angle of the equilibrium vector.

We know that equilibrium vector is equal in magnitude and in opposite direction to the given vector.

Therefore, magnitude of equilibrium vector=8 units

x-component of a vector= $v_x=vcos\theta$

Where v=Magnitude of vector

Using the formula

x-component of resultant vector= $v_x=8cos43=5.85$

y-component of resultant vector= $v_y=vsin\theta=8sin43=5.46$

x-component of equilibrium vector= $v_x=-5.85$

y-component of equilibrium vector= $-v_y=-5.46$

Because equilibrium vector lies in III quadrant

$\theta=tan^{-1}(\frac{v_x}{v_y})=tan^{-1}(\frac{-5.46}{-5.85})=43^{\circ}$

The angle $\theta'$ lies in III quadrant

In III quadrant ,angle = $\theta'+180^{\circ}$

Angle of equilibrium vector measured from positive x in counter clock wise direction=180+43=223 degree

4 0

1 year ago

What is the concentration of molecular oxygen (O2) in mol/L on a June day in Toronto when atmospheric pressure is 1.0 atm and th

saveliy_v [14]

Answer:

The concentration of mole evil at oxygen on that day is 0.00858 mol/L

Explanation:

Here, we want to calculate the concentration of molecular oxygen

The pressure on that day is 1.0 atm

Since oxygen is at a concentration of 21%, the pressure of oxygen will be 21/100 * 1 = 0.21 atm

Now let’s calculate the concentration;

From Ideal gas law;

PV = nRT

This can be written as;

P/RT = n/V

The term n/V refers to concentration;

Let’s make substitutions now;

P = pressure = 0.21 atm

R = molar gas constant = 0.0821 L•atm/mol•k

T = temperature = 25 = 25 + 273.15 = 298.15 K

Substituting these values, we have;

n/V = C = 0.21/(0.0821 * 298.15) = 0.00858 mol/L

6 0

2 years ago

If you are driving 72 km/h along a straight road and you look to the side for 4.0 s, how far do you travel during this inattenti

Ann [662]

We know that speed equals distance between time. Therefore to find the distance we have that d = V * t. Substituting the values d = (72 Km / h) * (1h / 3600s) * (4.0 s) = 0.08Km.Therefore during this inattentive period traveled a distance of 0.08Km

8 0

2 years ago

A rectangular coil of dimensions 5.40cm x 8.50cm consists of25 turns of wire. The coil carries a current of 15.0 mA.

Kazeer [188]

Answer:

(a) Magnetic moment will be $17.212\times 10^{-4}A-m^2$

(b) Torque will be $6.024\times 10^{-4}N-m$

Explanation:

We have given dimension of the rectangular 5.4 cm × 8.5 cm

So area of the rectangular coil $A=5.4\times 8.5=45.9cm^2=45.9\times 10^{-4}m^2$

Current is given as $i=15mA=15\times 10^{-3}A$

Number of turns N = 25

(A) We know that magnetic moment is given by $magnetic\ moment=NiA=25\times 45.9\times 10^{-4}\times 15\times 10^{-3}=17.212\times 10^{-4}A-m^2$

(b) Magnetic field is given as B = 0.350 T

We know that torque is given by $\tau =BINA=0.350\times 15\times 10^{-3}\times 25\times 45.9\times 10^{-4}=6.024\times 10^{-4}N-m$

4 0

1 year ago

A nonuniform beam 4.50 m long and weighing 1.40 kN makes an angle of 25.0° below the horizontal. It is held in position by a fri

liubo4ka [24]

Answer:

$T = 7.64 kN$

$F_y = 0.52 kN$ (Downwards)

$F_x = 3.23 kN$ (Towards Left)

Explanation:

As we know that beam is in equilibrium

So here we can use torque balance as well as force balance for the beam

Now by torque balance equation at the pivot we can say

$F(4.50 cos\theta) + mg(2cos\theta) = T \times 3$

As we know that

mg = 1.40 kN

F = 5 kN

so we will have

$5 kN(4.50 cos25) + 1.40 kN(2 cos25) = 3 T$

$T = 7.64 kN$

Now force balance in vertical direction

$F + mg = Tsin65 + F_y$

$5 + 1.40 = 7.64 sin65 + F_y$

$F_y = 0.52 kN$ (Downwards)

Force balance in horizontal direction

$F_x = T cos65$

$F_x = 7.64 cos65$

$F_x = 3.23 kN$ (Towards Left)

7 0

2 years ago