Question

A student pulls a sled of mass m = 82.0 kg with a force of F = 160N, and the force makes an angle of θ = 15 degrees with respect to the horizontal. Ignoring friction the normal force exerted on the sled by the ground is A) 649.05N;

B) 762.2N;
C) 527.90N;
D) 845.01N;

Answer 1

The normal force exerted on the sled by the ground is 762.2 N.

(B) is correct option.

Step-by-step explanation:

Given that,

Mass of sled m=82.0 kg

Force = 160 N

Angle = 15 degrees

We need to calculate the component of force

X- component,

`F_(x)=F\cos\theta`

Put the value into the formula

`F_(x)=160\cos15`

`F_(x)=154.54\ N`

Y- component,

`F_(y)=F\sin\theta`

Put the value into the formula

`F_(y)=160\sin15`

`F_(y)=41.41\ N`

We need to calculate the force due to gravity

Using formula of force

`F_(g) = mg`

`F_(g)=82.0*9.8`

`F_(g)=803.6\ N`

We need to calculate the normal force

Using formula of normal force

`F_(n)=F_(g)-F_(y)`

Put the value into the formula

`F_(n)=803.6-41.41`

`F_(n)=762.2\ N`

Hence, The normal force exerted on the sled by the ground is 762.2 N.

Answer 2

Answer:B

Step-by-step explanation:

Given

mass of sledge=82 kg

Force on sledge=160 N

degree
`=15^(\circ)`

force has two component sin and cos

Normal reaction
`=mg-F\sin \theta`

`N=82* 9.8-160\sin 15`

`N=803.6-41.41=762.19 N`